Sequences and Series of Functions MCQ Quiz - Objective Question with Answer for Sequences and Series of Functions - Download Free PDF

Last updated on Jul 1, 2025

Latest Sequences and Series of Functions MCQ Objective Questions

Sequences and Series of Functions Question 1:

For integers n ≥ 0, Let fn : [-1, 0] → ℝ be defined by

\(\rm f_n(x)=\frac{x}{(1-x)^n}\)

Which of the following statements is true about the series \(\rm \Sigma_{n=0}^\infty f_n\) ?

  1. The series is neither absolutely convergent nor uniformly convergent. 
  2. The series is both absolutely convergent and uniformly convergent. 
  3. The series is absolutely convergent but not uniformly convergent.  
  4. The series is uniformly convergent but not absolutely convergent. 

Answer (Detailed Solution Below)

Option 3 : The series is absolutely convergent but not uniformly convergent.  

Sequences and Series of Functions Question 1 Detailed Solution

Concept:

Absolute and Uniform Convergence of Function Series:

  • Given: \( f_n(x) = \dfrac{x}{(1 - x)^n} \), for all integers \( n \geq 0 \) and \( x \in [-1, 0] \).
  • We are asked to examine the convergence behavior of \( \sum_{n=0}^\infty f_n(x) \) on \( [-1, 0] \) — both in terms of absolute convergence and uniform convergence.
  • Absolute convergence: A function series \( \sum f_n(x) \) is absolutely convergent if \( \sum |f_n(x)| \) converges for all \( x \in [-1, 0] \).
  • Uniform convergence: A function series converges uniformly if \( \sup_{x \in [-1, 0]} |S_N(x) - S(x)| \to 0 \) as \( N \to \infty \).

 

Calculation:

Given,

\( f_n(x) = \dfrac{x}{(1 - x)^n} \)

Step 1: Absolute Convergence

For each fixed \( x \in [-1, 0] \), we examine \( \sum_{n=0}^\infty |f_n(x)| = \sum_{n=0}^\infty \left| \frac{x}{(1 - x)^n} \right| \)

Note:

  • For \( x = 0 \), we have \( f_n(0) = 0 \), so series trivially converges.
  • For \( x \in [-1, 0) \), \( 1 - x \geq 1 \Rightarrow (1 - x)^n \geq 1 \), and \( x < 0 \Rightarrow |f_n(x)| = \frac{-x}{(1 - x)^n} \).
  • The function \( |f_n(x)| \) decays exponentially fast, hence \( \sum |f_n(x)| \) converges for each \( x \in [-1, 0] \).

⇒ So the series is absolutely convergent on \( [-1, 0] \).

Step 2: Uniform Convergence

  • Uniform convergence of a sequence \( f_n(x) \to 0 \) means: \( \sup_{x \in [-1, 0]} |f_n(x)| \to 0 \) as \( n \to \infty \).

 

We analyze: \( \sup_{x \in [-1, 0]} |f_n(x)| = \sup_{x \in [-1, 0]} \left| \frac{x}{(1 - x)^n} \right| \).

Let us define: \( g_n(x) = \left| \frac{x}{(1 - x)^n} \right| \).

Since \( x \le 0 \), we write: \( g_n(x) = \frac{-x}{(1 - x)^n} \).

To check uniform convergence, examine: \( \sup_{x \in [-1, 0]} g_n(x) \).

Let’s maximize \( g_n(x) \) over \( [-1, 0] \).

Set: \( g_n(x) = \frac{-x}{(1 - x)^n} \), and take derivative with respect to \( x \) to find maximum.

Alternatively, test values:

At \( x = -1 \Rightarrow f_n(-1) = -\frac{1}{2^n} \)

At \( x = -\frac{1}{n} \), we get: \( |f_n(-\frac{1}{n})| = \frac{1/n}{(1 + \frac{1}{n})^n} = \frac{1}{n} \cdot \left( \frac{n}{n+1} \right)^n \approx \frac{1}{n e} \)

So \( \sup_{x \in [-1, 0]} |f_n(x)| \ge \frac{1}{n e} \not\to 0 \)

This implies: \( \sup_{x \in [-1, 0]} |f_n(x)| \not\to 0 \) as \( n \to \infty \)

∴ Since the supremum of \( |f_n(x)| \) does not tend to zero, the sequence \( f_n(x) \) is not uniformly convergent  on \( [-1, 0] \).

Sequences and Series of Functions Question 2:

Let f : [0, 1] → [1, ∞) be defined by \(\rm f(x)=\frac{1}{1-x}\), For n ≥ 1, let pn(x) = 1 + x + ....+ xn, Then which of the following statements are true? 

  1. f(x) is not uniformly continuous on [0, 1)
  2. The sequence (pn(x)) converges to f(x) pointwise on [0, 1) 
  3. The sequence (pn(x)) converges to f(x) uniformly on [0, 1) 
  4. The sequence (pn(x)) converges to f(x) uniformly on [0, c| for every 0 < c < 1 

Answer (Detailed Solution Below)

Option :

Sequences and Series of Functions Question 2 Detailed Solution

Concept:

Pointwise Convergence:

Let \(\{f_n(x)\}\) be a sequence of functions defined on a domain  D . The sequence \(\{f_n(x)\}\) is said to

converge pointwise to a function  \(f(x)\) on  D  if, for every point \(x \in D\)  and for every \(\epsilon >0\), there exists an integer  N

 such that for all  \(n \geq N\), the following condition holds

\(|f_n(x) - f(x)| < \epsilon.\)
 

In other words, for each fixed point \(x \in D\), the values of the functions \(f_n(x) \) get arbitrarily close to \(f(x)\) as \(n \to \infty \).

Uniform Convergence:

Let \(\{f_n(x)\}\) be a sequence of functions defined on a domain D. The sequence \(\{f_n(x)\}\) is said to converge uniformly to a function

 f(x) on  D if for every \(\epsilon >0\), there exists an integer N  such that for all \(n \geq N\) and for all \(x \in D\), the following condition holds

\(|f_n(x) - f(x)| < \epsilon.\)
 

Explanation:

\(f(x) = \frac{1}{1 - x} for x \in [0, 1)\)

For \(n \geq 1\) , let \(p_n(x) = 1 + x + x^2 + \dots + x^n \). This is the sum of the first n+1 terms of a geometric series.

We know that the infinite geometric series \(1 + x + x^2 + \dots = \frac{1}{1 - x} for |x| < 1\) , which means \(p_n(x)\) is an approximation of f(x) as \(n \to \infty\)  .

Option 1: The function \(f(x) = \frac{1}{1 - x}\) blows up as \(x \to 1\) , meaning that as we get closer to 1, the function

value grows without bound. This implies that f(x) is not uniformly continuous because the difference in

function values for close inputs near 1 can be arbitrarily large.

Option 1) is true.

Option 2: The sequence \(p_n(x) = 1 + x + \dots + x^n\) is a partial sum of the geometric series, and we know that

the infinite sum converges to \(f(x) = \frac{1}{1 - x}\). Therefore, \(p_n(x)\) converges to f(x) pointwise on [0, 1) .

Option 2 is true.

Option 3: Uniform convergence requires that the convergence happens uniformly across the entire interval.

Near x = 1 , the function f(x) grows very large, and the sequence \(p_n(x)\) does not converge uniformly

because the convergence slows down significantly near x = 1 .

Option 3 is false.

Option 4: This is true because for any c < 1 , the function \(f(x) = \frac{1}{1 - x}\) is continuous and well-behaved

on the interval [0, c] , and the sequence \(p_n(x)\)  will converge uniformly on compact subintervals away from 1.

Option 4 is true.

The correct options are Option 1), Option 2), and Option 4).

Sequences and Series of Functions Question 3:

Let (an)n≥1 be a bounded sequence of real numbers such that \(\rm \lim_{n\rightarrow \infty}a_n\) does not exist. Let S = {l ∈ ℝ : there exists a subsequence of (an) converges to l}.

Which of the following statements are necessarily true? 

  1. S is the empty set 
  2. S has exactly one element 
  3. S has at least two elements 
  4. S has to be a finite set 

Answer (Detailed Solution Below)

Option :

Sequences and Series of Functions Question 3 Detailed Solution

Concept:

Bounded Sequence: A bounded sequence means that all the elements of the sequence lie within some fixed

interval [-M, M] for some positive M. This ensures that the sequence does not diverge to infinity or negative infinity.

Non-existence of a Limit: The condition \(\limsup_{n \to \infty} a_n \neq \liminf_{n \to \infty} a_n\) implies that the sequence oscillates

between multiple values and does not converge to a single value.

Bolzano - Weierstrass Theorem: This theorem states that every bounded sequence has a convergent subsequence.

Therefore, there exist subsequences of \((a_n)\) that converge to different limit points.

Structure of Set S: The set S contains all the limit points of the subsequences of \((a_n)\).

Explanation:

Option 1: This cannot be true because, by the Bolzano - Weierstrass theorem, there must be at least one

convergent subsequence. Therefore, S contains at least one element.

False.

Option 2: This cannot be true because if the sequence does not converge, it must oscillate, which implies

that there are multiple limit points. Hence, S will contain more than one element.

False.

Option 3: Since the sequence does not converge to a single value, there must be at least two

distinct subsequential limits. Therefore, S must have at least two elements.

True.

Option 4: This is not necessarily true because a bounded sequence can have an infinite number

of subsequential limits. For example, the sequence could oscillate between an infinite set of values.

False.

The correct answer is Option 3).

Sequences and Series of Functions Question 4:

For each n ≥ 1 define fn : ℝ → ℝ by \(\rm f_n(x)=\frac{x^2}{√{x^2+\frac{1}{n}}}, \) x ∈ ℝ

where √ denotes the non-negative square root. Wherever \(\rm \lim_{n \rightarrow \infty}f_n(x)\) exists, denote it by f(x). Which of the following statements is true? 

  1. There exists x ∈ ℝ such that f(x) is not defined 
  2. f(x) = 0 for all x ∈ ℝ 
  3. f(x) = x for all x ∈ ℝ 
  4. f(x) = |x| for all x ∈ ℝ

Answer (Detailed Solution Below)

Option 4 : f(x) = |x| for all x ∈ ℝ

Sequences and Series of Functions Question 4 Detailed Solution

Concept:

Limit of a Sequence of Functions:

1. Let \(\{f_n\}\) be a sequence of functions defined on a set D . We say that \(f_n\) converges pointwise to a function

 f  on D if, for every x \(\in\) D

\(\lim_{n \to \infty} f_n(x) = f(x).\)

2. A stronger form of convergence is uniform convergence. The sequence \(\{f_n\}\) converges uniformly to a function  f  on D if

\(\lim_{n \to \infty} \sup_{x \in D} |f_n(x) - f(x)| = 0.\)

Explanation: The problem gives a sequence of functions\( f_n: \mathbb{R} \to \mathbb{R}\) defined by

\(f_n(x) = \frac{x^2}{\sqrt{x^2 + \frac{1}{n}}}\)

and asks about the limit of \(f_n(x) \) as \(n \to \infty \), denoted by \( f(x)\) . We are tasked with determining which statement about f(x) is true.

We are asked to take the limit \(n \to \infty \) of the function:

 \(f_n(x) = \frac{x^2}{\sqrt{x^2 + \frac{1}{n}}} \)   

As \(n \to \infty \), the term \(\frac{1}{n} \to 0 \). So, for large n , the function \(f_n(x) \) approaches

 \(lim_{n \to \infty} f_n(x) = \frac{x^2}{\sqrt{x^2}} = \frac{x^2}{|x|}\)
 

Case 1: \(x \neq 0\)

For \(x \neq 0\) we have, \( f(x) = \frac{x^2}{|x|} = |x|\)
 

Case 2: \(x =0\)
When \(x =0\) , the function becomes \(f_n(0) = \frac{0^2}{\sqrt{0^2 + \frac{1}{n}}} = 0\)

Therefore, as \(n \to \infty \), we get f(0) = 0 .

The function f(x) , \(n \to \infty \) , is given by

   
  \( f(x) = \begin{cases} |x|, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases} \)

This function is equal to |x| for all \(x \in \mathbb{R} \)

Therefore, The correct option is 4).

Sequences and Series of Functions Question 5:

Let (fn)n≥1 be the sequence of functions defined on [0, 1] by

\(f_n(x)=x^n \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right) .\)

Which of the following statements are true? 

  1. (fn) converges pointwise on [0, 1].
  2. (fn) converges uniformly on compact subsets of [0,1) but not on [0,1).
  3. (fn) converges uniformly on [0,1) but not on [0,1].
  4. (fn) converges uniformly on [0, 1].

Answer (Detailed Solution Below)

Option :

Sequences and Series of Functions Question 5 Detailed Solution

Concept:

Pointwise convergent: A sequence of functions f1, f2, … , fn, … : E → ℝ (E is a subset of ℝ) is said to converge pointwise on E to function f: E → ℝ if and only if

f(x) = \(\lim_{n\to \infty}f_n(x)\) for all x ∈ E 

Uniform convergent: Given a sequence of functions fn: E → ℝ, we say fn converges uniformly to f if and only if \(\lim_{n\to \infty}\left(\sup_{x∈ E}|f_n(x)-f(x)|\right)\) = 0

Explanation:

\(f_n(x)=x^n \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right) .\)

(1): For x = 0

f(x) = \(\lim_{n\to \infty}f_n(x)\) = \(\lim_{n\to \infty}\)\(x^n \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right)\) = 0

For x = 1                                                  

f(x) = \(\lim_{n\to \infty}\)\(x^n \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right)\) = \(\lim_{n\to \infty}\)\(1^n \log \left(\frac{4+{1^{\frac{1}{4}}}}{5}\right)\) = log 1 = 0

For 0 < x < 1,

f(x) = \(\lim_{n\to \infty}\)\(x^n \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right)\) = 0

So, (fn) converges pointwise on [0, 1].

(1) is correct.

(4): fn(x) = \(x^n \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right)\) is a sequence of continuous functions converges pointwise to a continuous function f(x) = 0 on a compact set [0, 1]

xn+1 ≤ xn ∀ x ∈ [0, 1]

\(x^{n+1} \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right)\) ≤ \(x^n \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right)\) ∀ x ∈ [0, 1]

fn+1(x) ≤ fn(x) ∀ x ∈ [0, 1]

So, {fn(x)} is a increasing sequence

hence (fn) converges uniformly in [0, 1]

(4) is correct

Top Sequences and Series of Functions MCQ Objective Questions

For each n ≥ 1 define fn : ℝ → ℝ by \(\rm f_n(x)=\frac{x^2}{√{x^2+\frac{1}{n}}}, \) x ∈ ℝ

where √ denotes the non-negative square root. Wherever \(\rm \lim_{n \rightarrow \infty}f_n(x)\) exists, denote it by f(x). Which of the following statements is true? 

  1. There exists x ∈ ℝ such that f(x) is not defined 
  2. f(x) = 0 for all x ∈ ℝ 
  3. f(x) = x for all x ∈ ℝ 
  4. f(x) = |x| for all x ∈ ℝ

Answer (Detailed Solution Below)

Option 4 : f(x) = |x| for all x ∈ ℝ

Sequences and Series of Functions Question 6 Detailed Solution

Download Solution PDF

Concept:

Limit of a Sequence of Functions:

1. Let \(\{f_n\}\) be a sequence of functions defined on a set D . We say that \(f_n\) converges pointwise to a function

 f  on D if, for every x \(\in\) D

\(\lim_{n \to \infty} f_n(x) = f(x).\)

2. A stronger form of convergence is uniform convergence. The sequence \(\{f_n\}\) converges uniformly to a function  f  on D if

\(\lim_{n \to \infty} \sup_{x \in D} |f_n(x) - f(x)| = 0.\)

Explanation: The problem gives a sequence of functions\( f_n: \mathbb{R} \to \mathbb{R}\) defined by

\(f_n(x) = \frac{x^2}{\sqrt{x^2 + \frac{1}{n}}}\)

and asks about the limit of \(f_n(x) \) as \(n \to \infty \), denoted by \( f(x)\) . We are tasked with determining which statement about f(x) is true.

We are asked to take the limit \(n \to \infty \) of the function:

 \(f_n(x) = \frac{x^2}{\sqrt{x^2 + \frac{1}{n}}} \)   

As \(n \to \infty \), the term \(\frac{1}{n} \to 0 \). So, for large n , the function \(f_n(x) \) approaches

 \(lim_{n \to \infty} f_n(x) = \frac{x^2}{\sqrt{x^2}} = \frac{x^2}{|x|}\)
 

Case 1: \(x \neq 0\)

For \(x \neq 0\) we have, \( f(x) = \frac{x^2}{|x|} = |x|\)
 

Case 2: \(x =0\)
When \(x =0\) , the function becomes \(f_n(0) = \frac{0^2}{\sqrt{0^2 + \frac{1}{n}}} = 0\)

Therefore, as \(n \to \infty \), we get f(0) = 0 .

The function f(x) , \(n \to \infty \) , is given by

   
  \( f(x) = \begin{cases} |x|, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases} \)

This function is equal to |x| for all \(x \in \mathbb{R} \)

Therefore, The correct option is 4).

Sequences and Series of Functions Question 7:

For each n ≥ 1 define fn : ℝ → ℝ by \(\rm f_n(x)=\frac{x^2}{√{x^2+\frac{1}{n}}}, \) x ∈ ℝ

where √ denotes the non-negative square root. Wherever \(\rm \lim_{n \rightarrow \infty}f_n(x)\) exists, denote it by f(x). Which of the following statements is true? 

  1. There exists x ∈ ℝ such that f(x) is not defined 
  2. f(x) = 0 for all x ∈ ℝ 
  3. f(x) = x for all x ∈ ℝ 
  4. f(x) = |x| for all x ∈ ℝ

Answer (Detailed Solution Below)

Option 4 : f(x) = |x| for all x ∈ ℝ

Sequences and Series of Functions Question 7 Detailed Solution

Concept:

Limit of a Sequence of Functions:

1. Let \(\{f_n\}\) be a sequence of functions defined on a set D . We say that \(f_n\) converges pointwise to a function

 f  on D if, for every x \(\in\) D

\(\lim_{n \to \infty} f_n(x) = f(x).\)

2. A stronger form of convergence is uniform convergence. The sequence \(\{f_n\}\) converges uniformly to a function  f  on D if

\(\lim_{n \to \infty} \sup_{x \in D} |f_n(x) - f(x)| = 0.\)

Explanation: The problem gives a sequence of functions\( f_n: \mathbb{R} \to \mathbb{R}\) defined by

\(f_n(x) = \frac{x^2}{\sqrt{x^2 + \frac{1}{n}}}\)

and asks about the limit of \(f_n(x) \) as \(n \to \infty \), denoted by \( f(x)\) . We are tasked with determining which statement about f(x) is true.

We are asked to take the limit \(n \to \infty \) of the function:

 \(f_n(x) = \frac{x^2}{\sqrt{x^2 + \frac{1}{n}}} \)   

As \(n \to \infty \), the term \(\frac{1}{n} \to 0 \). So, for large n , the function \(f_n(x) \) approaches

 \(lim_{n \to \infty} f_n(x) = \frac{x^2}{\sqrt{x^2}} = \frac{x^2}{|x|}\)
 

Case 1: \(x \neq 0\)

For \(x \neq 0\) we have, \( f(x) = \frac{x^2}{|x|} = |x|\)
 

Case 2: \(x =0\)
When \(x =0\) , the function becomes \(f_n(0) = \frac{0^2}{\sqrt{0^2 + \frac{1}{n}}} = 0\)

Therefore, as \(n \to \infty \), we get f(0) = 0 .

The function f(x) , \(n \to \infty \) , is given by

   
  \( f(x) = \begin{cases} |x|, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases} \)

This function is equal to |x| for all \(x \in \mathbb{R} \)

Therefore, The correct option is 4).

Sequences and Series of Functions Question 8:

For a positive integer n, let f(n) denote the nth derivative of f. Suppose an entire function f satisfies f(2) + f = 0.

Which of the following is correct?

  1. (f(n)(0))n≥1 is convergent.
  2. limn→∞ f(n)(0) = 1.
  3. limn→∞f(n)(0) = -1.
  4. (|f(n) (0)|)n≥1 has a convergent subsequence.

Answer (Detailed Solution Below)

Option 4 : (|f(n) (0)|)n≥1 has a convergent subsequence.

Sequences and Series of Functions Question 8 Detailed Solution

Explanation:

Let f(z) = sin z

Then f1(z) = cos z and f2(z) = -sin z

So, f(2) + f = 0 satisfy

Also, f3(z) = - cos z, f4(z) = sin z, f5(z) = cos z ...

Hence we get

f1(0) = 1, f2(0) = 0, f3(0) = -1, f4(0) = 0, f5(0) = 1, ....

Therefore we got the sequence 

{f(n)(0)} = {1, 0, -1, 0, 1, ...}

So, f(n)(0) has 3 limit points 1, 0, -1

Hence (f(n)(0))n≥1 is not convergent

(1) is false.

Also (2), (3) false

and (4) is true

Sequences and Series of Functions Question 9:

The sequence of real-valued function

fn(x) = xn, x ∈ [0, 1] ∪ {2} is

  1. point wise convergent
  2. uniformly convergent
  3. does not uniformly convergent
  4. point wise limit is f(x) = \(\left\{ \begin{matrix} 0, x = 0 \\\ 1, x \ne 0 \end{matrix} \right.\)

Answer (Detailed Solution Below)

Option 3 : does not uniformly convergent

Sequences and Series of Functions Question 9 Detailed Solution

Concept:

(i) Pointwise convergent: A sequence of functions fn: E → ℝ (where E is a subset of ℝ is said to be converges pointwise on E to function f: E → ℝ if and only if limn→∞fn(x) = f(x) for all x ∈ E

(ii) Uniformly convergent: a sequence of function fn: E → ℝ is said to be converges uniformly to f if and only if limn→∞ \(\sup_{x\in E}|f_n(x)-f(x)|\) = 0

(iii) If a sequence is not pointwise convergent then sequence is not uniformly convergent.
Explanation:

Let α ∈ [0, 1] ∪ {2}, then

Case I: If α = 0, then

f(x) = \(\lim_{n \rightarrow \infty} f_n(α)\)

\(\lim_{n \rightarrow \infty} f_n(0)\)

\(\lim_{n \rightarrow \infty} 0 = 0\)

Case II: If α = 1, then

f(x) = \(\lim_{n \rightarrow \infty} f_n(α)\) = \(\lim_{n \rightarrow \infty} f_n(1)\)

\(\lim_{n \rightarrow \infty} 1 = 1\)

Case III: If 0 < α < 1, then

f(α) = \(\lim_{n \rightarrow \infty} f_n(α)\)

\(\lim_{n \rightarrow \infty}a^n\)

= 0

Case IV: If α = 2, then,

f(x) = \(\lim_{n \rightarrow \infty} f_n ( \alpha)\)

\(\lim_{n \rightarrow \infty} f_n ( 2)\)

\(\lim_{n \rightarrow \infty}2^n\)

= does not exist

Hence, fn(x) is not point-wise convergent.

⇒ fn(x) is not uniformly convergent.

Option (3) is correct

Sequences and Series of Functions Question 10:

Which of the following are true?

  1. For n ≥ 1, the sequence of functions fn : (0, 1) → (0, 1) defined by fn(x) = xis uniformly convergent.
  2. For n ≥ 1, the sequence of functions fn : (0, 1) → (0, 1) defined by fn(x) = \(\rm\frac{x^n}{\log (n+1)}\) is uniformly convergent.
  3. For n ≥ 1, the sequence of functions fn : (0, 1) → (0, 1) defined by fn(x) = \(\rm\frac{x^n}{1+x^n}\) is uniformly convergent.
  4. For n ≥ 1, the sequence of functions f: (0, 1) → (0, 1) defined by fn(x) = \(\rm\frac{x^n}{1+n x^n}\) is not uniformly convergent.

Answer (Detailed Solution Below)

Option :

Sequences and Series of Functions Question 10 Detailed Solution

Concept:

Pointwise convergent: A sequence of functions f1, f2, … , fn, … : E → ℝ (where E is a subset of ℝ is said to be converges pointwise on E to function f: E → ℝ if and only if \(\lim_{n\to\infty}f_n(x)=f(x)\) ∀ x ∈ E

Uniform convergent: Given a sequence of function fn: E → ℝ, we say fn converges uniformly to f if and only if \(\lim_{n\to\infty}(\sup_{x\in E}|f_n(x)-f(x)|)\) = 0 this is called Mn test

Result: If the pointwise limit of a sequence of function is not continuous, then the sequence of function is not uniformly continuous. 

Explanation:

(1): fn(x) = xn, x ∈ (0, 1)

So f(x) = \(\lim_{n\to\infty}x^n \) = 0 for x ∈ (0, 1)

for x = \(1-\frac1n\)

\(\lim_{n\to\infty}f_n(x)\)  = \(\lim_{n\to\infty}(1-\frac1n)^n\) = e-1 does not tends to 0  

So  fn(x) = xn, x ∈ (0, 1) is not uniformly convergent.

Option (1) is false

(2): fn(x) = \(\rm\frac{x^n}{\log (n+1)}\) 

f(x) = 0 for x ∈ (0, 1)

Option (2) is true.

(3): fn(x) = \(\rm\frac{x^n}{1+x^n}\) 

So f(x) = \(\lim_{n\to\infty} \frac{x^n}{1+x^n} \) = 0 for x ∈ (0, 1)

for x = \(1-\frac1n\)

\(\lim_{n\to\infty}f_n(x)\)  = \(\lim_{n\to\infty}\frac{(1-\frac1n)^n}{1+(1-\frac1n)^n} = \frac{e^{-1}}{1+e^{-1}} \neq 0\)  does not tends to 0  

So fn(x) = \(\rm\frac{x^n}{1+x^n}\), x ∈ (0, 1) is not uniformly convergent.

Option (3) is false.

(4): fn(x) = \(\rm\frac{x^n}{1+n x^n}\)

So f(x) = \(\lim_{n\to\infty} \frac{x^n}{1+nx^n} \) = 0 for x ∈ (0, 1)

And this holds for every x. Hence the  fn(x) = \(\rm\frac{x^n}{1+n x^n}\), x ∈ (0, 1) is uniformly convergent.

Option (4) is false.

Sequences and Series of Functions Question 11:

The sequence

\( f_n(x)= \begin{cases}n^2 x, & 0 \leq x \leq \frac{1}{n} \\ -n^2 x+2 n, & \frac{1}{n} \leq x \leq \frac{2}{n} \\ 0, & \frac{2}{n} \leq x \leq 1\end{cases} \)

  1. not pointwise convergent
  2. the point wise limit is f(x) = x
  3. \( \lim _{n \rightarrow \infty} \int_0^1 f_n(x) d x=\int_0^1 f(x) d x\)
  4. \(\lim _{n \rightarrow \infty} \int_0^1 f_n(x) d x \neq \int_0^1 f(x) d x\)

Answer (Detailed Solution Below)

Option 1 : not pointwise convergent

Sequences and Series of Functions Question 11 Detailed Solution

Concept:

(i) Pointwise convergent: A sequence of functions fn: E → ℝ (where E is a subset of ℝ is said to be converges pointwise on E to function f: E → ℝ if and only if limn→∞fn(x) = f(x) for all x ∈ E

(ii) Uniformly convergent: a sequence of function fn: E → ℝ is said to be converges uniformly to f if and only if limn→∞ \(\sup_{x\in E}|f_n(x)-f(x)|\) = 0

Explanation:

\( f_n(x)= \begin{cases}n^2 x, & 0 ≤ x ≤ \frac{1}{n} \\ -n^2 x+2 n, & \frac{1}{n} ≤ x ≤ \frac{2}{n} \\ 0, & \frac{2}{n} ≤ x ≤ 1\end{cases} \)

For 0 ≤ x ≤ 1/n

limn→∞fn(x) = limn→∞ n2x = ∞

For 1/n ≤ x ≤ 2/n

limn→∞fn(x) = limn→∞ (- n2x + 2n) = ∞

For 2/n ≤ x ≤ 1

limn→∞fn(x) = limn→∞ 0 = 0

So here we can see that fn(x) does not converges to a single limit as n tends to infinite.

Hence fn(x) not pointwise convergent

Option (1) is true

Sequences and Series of Functions Question 12:

Let \(\rm f(x)=\frac{\log (2+x)}{\sqrt{1+x}}\) for x ≥ 0, and \(\rm a_m=\frac{1}{m} \int_0^m f(t) d t\) for every positive integer m. Then the sequence \(\rm \left\{a_m\right\}_{m=1}^{\infty}\)

  1. diverges to + ∞.
  2. has more than one limit point.
  3. Converges and satisfies \(\rm \displaystyle\lim _{n \rightarrow \infty} a_m=\frac{1}{2} \log 2 \)
  4. Converges and satisfies \(\rm \displaystyle \lim _{n \rightarrow \infty} a_m=0\)

Answer (Detailed Solution Below)

Option 4 : Converges and satisfies \(\rm \displaystyle \lim _{n \rightarrow \infty} a_m=0\)

Sequences and Series of Functions Question 12 Detailed Solution

Explanation:

\(\int_0^m f(t) d t\) = \(\int_0^m \frac{\log (2+t)}{\sqrt{1+t}} d t\)

let 1 + t = z2 ⇒ dt = 2zdz  

∴ \(\int_0^m f(t) d t\) = \(2\int_1^{\sqrt{1+m}} \log(1+z^2) d z\)

                  = 2\(\left[z\log(1+z^2)\right]_1^{\sqrt{1+m}}-\int_1^{\sqrt{1+m}}\frac{2z^2}{1+z^2}dz\)

                   = 2\(\left[z\log(1+z^2)\right]_1^{\sqrt{1+m}}-2\left[z-tan^{-1}z\right]_1^{\sqrt{1+m}}\)

                  = 2\({\sqrt{1+m}}\log(t+2)\) - 2 log 2 - 4 \({\sqrt{1+m}}\) + 4 + 4\(\tan^{-1}{\sqrt{1+m}}\) - 4\(\frac{π}{4}\)

                   = 2\({\sqrt{1+m}}\log(t+2)\) - 2 log 2 - 4 \({\sqrt{1+m}}\)  + 4\(\tan^{-1}{\sqrt{1+m}}\) + 4 - π 

Therefore 

\(\rm a_m=\frac{1}{m} \int_0^m f(t) d t\) = 2\(​​\frac{\sqrt{1+m}}{m}\) log(t +2) - 2log 2 + 4 - π - 4 \(​​\frac{\sqrt{1+m}}{m}\) +\(\frac4m\)\(\tan^{-1}{\sqrt{1+m}}\) - \(\frac{4-\pi}{m}\)   

Hence \(\rm \displaystyle \lim _{n \rightarrow \infty} a_m=0\)

Option (4) is corrcet

Sequences and Series of Functions Question 13:

Consider the following statements:
(a) Let f be a continuous function on [1, ∞) taking non-negative values such that \(\displaystyle\int_1^{\infty}\) f(x) dx converges. Then \(\rm\displaystyle\sum_{n \geq 1}\) f(n) converges.

(b) Let f be a function on [1, ∞) taking non-negative values such that \(\displaystyle\int_1^{\infty}\) f(x) dx converges. Then \(\rm\displaystyle\lim _{x \rightarrow \infty}\) f(x) = 0.

(c) Let f be a continuous, decreasing function on [1, ∞) taking non-negative values such that \(\displaystyle\int_1^{\infty}\) f(x) dx does not converge. Then \(\rm\displaystyle\sum_{n \geq 1}\) f(n) does not converge.

Which of the following options are true?

  1. (a), (b) and (c) all are true.
  2. Both (a) and (b) are false.
  3. (c) is true.
  4. (b) is true.

Answer (Detailed Solution Below)

Option :

Sequences and Series of Functions Question 13 Detailed Solution

Concept:

Let N be a natural number (non-negative number), and it is a monotonically decreasing function, then the function is defined as f : [N,∞ ]→ ℝ Then the series \(\sum_{m=N}^{\infty}f(m)\) and \(\int_N^{\infty}f(t)dt\)

converges or diverges together

Solution:

For (1) -

Consider the function \(f(x) = \frac{1}{x^2}\) for \(x \geq 1\). This function is continuous on\( [1, \infty) \)  and takes non-negative values. The integral of f(x) from 1 to infinity converges:

\(\int_1^\infty \frac{1}{x^2} dx\)  = \(\left[-\frac{1}{x}\right]_1^\infty = 1\)

However, if we evaluate f(n) for integers n, we get:

\(f(n) = \frac{1}{n^2}\)

The series formed by summing these terms does not converge:

\(\sum_{n=1}^\infty \frac{1}{n^2}\)

This is the classic example of the Basel problem, which converges to \(\frac{\pi^2}{6}\).

Therefore, the given statement does not hold in general

For (2) -

The fact that \(\int_1^\infty f(x) \, dx\) converges does not necessarily imply that f(x) = 0  for all x in \([1, \infty).\)

However, it does imply that f(x) = 0 almost everywhere on \([1, \infty).\)

For instance, take a function f(x) that's zero everywhere except on a set of points in \([1, \infty)\) that has a total measure of zero.

This function would integrate to zero over \([1, \infty)\) despite not being identically zero.

Therefore, the condition that \(\int_1^{\infty} f(x)dx\) converges indeed implies that f(x) = 0 almost everywhere on \([1, \infty)\) but not necessarily for all points x in that interval.

Hence the statement is false.

(3): By using Cauchy Integral test 

\(f:[1,\infty) \rightarrow R\)  

f(x) > 0 for all x belongs  to R 

f is monotonic decreasing 

then \(\int_1^\infty f(x) dx\) and \(\rm\displaystyle\sum_{n \geq 1}\)f(x) converges or diverges together 

Therefore, By Cauchy integral test (c) is correct. 

Hence Option (2) & (3) are correct

Sequences and Series of Functions Question 14:

A sequence {fn} defined on set S is said to be uniformly bounded on S if

  1. there exists k < 0 such that |fn(x)| < k; for all x belongs to S and n belongs to \(\mathbb N\).
  2. there exists k > 0 such that |fn(x)| > k; for all x belongs to S and n belongs to \(\mathbb N\).
  3. there exists k > 0 such that |fn(x)| < k; for all x belongs to S and n belongs to \(\mathbb N\).
  4. there exists k < 0 such that |fn(x)| > k; for all x belongs to S and n belongs to \(\mathbb N\)

Answer (Detailed Solution Below)

Option 3 : there exists k > 0 such that |fn(x)| < k; for all x belongs to S and n belongs to \(\mathbb N\).

Sequences and Series of Functions Question 14 Detailed Solution

Explanation:

A sequence {fn} defined on set S is said to be uniformly bounded on S if there exists k > 0 such that |fn(x)| < k; for all x belongs to S and n belongs to \(\mathbb N\).

(3) is correct

Sequences and Series of Functions Question 15:

Let fn(x) = xn, ∀ x ∈ [0, 1] and for all positive integers n. Then,

  1. < fn > does not converge at any point
  2. < fn > converges at some points, but does not converge at some other points
  3. < fn > converges uniformly
  4. < fn > converges at every point, but does not converge uniformly

Answer (Detailed Solution Below)

Option 4 : < fn > converges at every point, but does not converge uniformly

Sequences and Series of Functions Question 15 Detailed Solution

Concept:

(i) Pointwise convergent: A sequence of functions fn: E → ℝ (where E is a subset of ℝ is said to be converges pointwise on E to function f: E → ℝ if and only if limn→∞fn(x) = f(x) for all x ∈ E

(ii) Uniformly convergent: a sequence of function fn: E → ℝ is said to be converges uniformly to f if and only if limn→∞ \(\sup_{x\in E}|f_n(x)-f(x)|\) = 0

Explanation:

We have, fn(x) = xn, ∀ x ∈ [0, 1]

⇒ \(f(x) = \left\{ \begin{matrix} 0, &0 \le x < 1 \\\ 1, &x = 1 \end{matrix} \right.\)

 \( ∴ M_n=\operatorname{Sup}_{x \in[0,1]}\left\{\left|f_n(x)-f(x)\right|\right\} \)

\(=\operatorname{Sup}_{x \in[0,1]}\left\{\left| x^n\right|\right\} = 1\)

∴ \(\lim_{n \rightarrow \infty}M_n = 1\)

⇒ fn(x) is not uniformly convergent but convergent at every point.

Option (4) is true

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