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

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 \( [-1, 0] \).

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).

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).

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).

Concept:

Pointwise convergent: A sequence of functions f₁, f₂, … , f_n, … : 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 f_n: 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): f_n(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]

f_n+1(x) ≤ f_n(x) ∀ x ∈ [0, 1]

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

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

(4) is correct

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).

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).

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

Concept:

(i) Pointwise convergent: A sequence of functions f_n: 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→∞f_n(x) = f(x) for all x ∈ E

(ii) Uniformly convergent: a sequence of function f_n: E → ℝ is said to be converges uniformly to f if and only if lim_n→∞ \(\sup_{x\in E}|f_n(x)-f(x)|\) = 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, f_n(x) is not point-wise convergent.

⇒ f_n(x) is not uniformly convergent.

Option (3) is correct

Concept:

Pointwise convergent: A sequence of functions f₁, f₂, … , f_n, … : 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 f_n: 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 M_n 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.

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

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

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

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

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]

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

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

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

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

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

Option (4) is true