Sequences & Series (Convergence) MCQ Quiz - Objective Question with Answer for Sequences & Series (Convergence) - Download Free PDF

Concept:

Understanding Sequences and Their Limits:

• Sequence: An ordered list of numbers defined by a function on natural numbers.
• Convergence: A sequence converges if it approaches a unique real number as .
• lim sup: The greatest accumulation point (limit of suprema of tails).
• lim inf: The smallest accumulation point (limit of infima of tails).
• Max sequence: is non-decreasing.
• Min sequence: is non-increasing.

Calculation:

Given,

⇒ For even :

⇒ For odd :

⇒ So the sequence oscillates between numbers close to +1 and -1, hence:

does not exist 

Option 1: " does not converge"

⇒ Correct.

Since the sequence oscillates between two values without settling on one, it diverges.

Option 2: ""

→ False.

Since 1 \), and this is the max forever,

⇒ not equal to 1

Option 3: ""

→ False.

which is minimum forever ⇒ for all large

So

Option 4: ""

→ False.

As shown,

Hence, not equal

∴ Only correct statement is Option 1.

Concept:

Behaviour of Two Sequences:

• Define .
• Monotone ↑ and Unbounded ↑ because 1 \) and .
• Define .
• Simplify ; it is bounded and tends to .

Calculation:

Given,

⇒

⇒ 1 \)

⇒ strictly ↑

⇒ For any ,  x \)

⇒   ⇒  for all 

">⇒ Cannot exceed any e \);  cannot go below any 

"> ∴ Only statement 1 is true; for every  there exists an  with x \).

Concept:

Power Series and Radius of Convergence:

• A power series is an infinite series of the form:
• The radius of convergence (R) is the distance from the origin within which the series converges.
• To find R, we use the root test or ratio test.
• Root Test Formula:
• Important Note: If the radius is R, the series converges absolutely for all .

Calculation:

Given,

General term:

Let us apply the root test:

⇒

⇒  

Now consider only the dominant factor:

⇒ as

⇒

⇒ So, the radius of convergence is:

Therefore, the series converges for all   

⇒ Option 1 , 2 and 3 are not correct 

⇒ Option 4 is correct 

∴ The correct statement is: The series converges for all x with  .

Explanation:

f(x) = 3/((1-x)(1+2x)) = A/(1-x) + B/(1+2x)

Solving for A and B, we get:

A = 1, B = 2

Therefore,  f(x) = 1/(1-x) + 2/(1+2x)

Now, we can use the geometric series expansion:

1/(1-x) = 1 + x + x² + x³ + x⁴ + ⋯ 

and 1/(1+2x) = 1 - 2x + 4x² - 8x³ + 16x⁴ + ⋯ 

Multiplying the second series by 2:

2/(1+2x) = 2 - 4x + 8x² - 16x³ + 32x⁴ + ⋯  

Now, we can add the two series to get the expansion of f(x):

f(x) = (1 + x + x² + x³ + x⁴ + ⋯ ) + (2 - 4x + 8x² - 16x³ + 32x⁴ + ⋯ )

To find the coefficient of x²:

x2 from the first series +x2 from the second series = +8x2

Therefore, the coefficient of x2 in the expansion of f(x) is 9

Hence 9 is the correct answer.

Explanation:

If    

Now   and   

     

So, (1) is true and (2) is false

Consider the function:

Then    but f'(x) does not exist at x = 1, so (3) is false. 

Consider    

Now   

But    Converges, so    Converges by comparison test

So,    Converges absolutely   converges to f(x) for each   


 
f(0) = 0

Also, f(x) = 0,   

Hence Option(1) and Option(4) are correct 

Concept:

Leibniz's test: A series of the form (-1)ⁿbn, where either all bn are positive or all bn are negative is convergent if

(i) |bn| decreases monotonically i.e., |bn+1| ≤ |bn|

(ii) 

Explanation:

an = (−1)n+1

    = (−1)n+1   

   = (−1)n+1

So series is  

So here b_n = , bn+1 = 

 so bn+1 n  

Also  =   = 0

Hence by Leibnitz's test  an is convergent.

Now the series is  =   =  

Hence by Limit comparison Test, it is divergent series by P - Test.

Hence the given series is conditionally convergent.

Option (3) is correct.

In Official answer key - Options (2) & (3) both are correct.

Explanation:

Tips: Try to discard options by taking suitable choice for n >.

option (1). Let a_n = 1 then (-1)ⁿ . ≠ 0

option (2). Let an = and  = then

  Not convergent.

option (3), option (4): (NB: may be you have to try with more) then one sea, n>

Let a_n = (-1)ⁿ then

But here fixed 'b S.t above series become cgt. You may take b = ½ or = -½ but not both otherwise uniqueness will be lost.

⇒ option (3) is false. 

option (4): As discussed earlier. take b = ½ and  = then about series becomes convergent. Hence option (4) is true.

Concept:

Supremum (sup): The supremum of a set is the least upper bound. For a sequence ,

  is the smallest number that is greater than or equal to all the terms of the sequence.

Infimum (inf): The infimum is the greatest lower bound. For a sequence ,  is the

largest number that is less than or equal to all the terms of the sequence.

Explanation:

Option 1:

The infimum and supremum of the sequence refer to its lower and upper bounds. If these two limits

coincide, it implies that the sequence is squeezing towards a single point.

This is a true statement, as if the infimum and supremum converge to the same point,

the sequence must converge to that point.

Option 2:

If the infimum of the sequence is equal to the limit of the sequence as , this implies that

the sequence stabilizes at this value, suggesting that it is converging to that point.

This is a true statement, as the sequence is converging to its infimum, implying that it has a limit.

Option 3:

Counter example:

Consider the sequence .

1. As , .

2. The supremum of the sequence is 

This sequence is clearly not constant because the values of an" id="MathJax-Element-145-Frame" role="presentation" style="position: relative;" tabindex="0">anan" id="MathJax-Element-69-Frame" role="presentation" style="position: relative;" tabindex="0">anan" id="MathJax-Element-163-Frame" role="presentation" style="position: relative;" tabindex="0">anan $a_n$ decrease as n" id="MathJax-Element-146-Frame" role="presentation" style="position: relative;" tabindex="0">nn" id="MathJax-Element-70-Frame" role="presentation" style="position: relative;" tabindex="0">nn" id="MathJax-Element-164-Frame" role="presentation" style="position: relative;" tabindex="0">nn n n $n$ increases. However, we still have:

This shows that  is not constant.

Option 4:

The supremum of a sequence is the least upper bound of the values in the sequence.

It is the smallest number that is greater than or equal to every term in the sequence.
  
The infimum of a sequence is the greatest lower bound of the values in the sequence.

It is the largest number that is less than or equal to every term in the sequence.

Now, if , this means that the least upper bound and the

greatest lower bound are the same. Let’s call this common value C.

Since the supremum C is an upper bound of the sequence, all terms in the sequence must be less than or equal to C.
Since the infimum C is a lower bound of the sequence, all terms in the sequence must be greater than or equal to C.

Therefore, for all n, the term  must satisfy , which implies that  = C for all n.

Hence, the required option is 3).an" id="MathJax-Element-145-Frame" role="presentation" style=" word-spacing: 0px; position: relative;" tabindex="0">" id="MathJax-Element-71-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-165-Frame" role="presentation" style="position: relative;" tabindex="0">

{an|n≥1}=limn→∞ansupc

$$\boxed{\text{Option 3}}.$$

Concept:

(i) Every convergent sequence is bounded.

Explanation:

{xn} is a convergent sequence in ℝ. So it is bounded.

Then there exists a real number M such that |x_n| ≤ M.

 {yn} is a bounded sequence in ℝ

Then there exists a real number L such that |yn| ≤ L.

Now, |xn + yn| ≤ |xn| + |yn| ≤ M + L

So, {xn + yn} is bounded.

Option (2) is true.

Let {xn} = {} and {yn} = {(-1)ⁿ} then {xn} is a convergent sequence in ℝ and {yn} is a bounded sequence in ℝ.

But {xn + yn} = { + (-1)n} which is not convergent and it has convergent and bounded subsequence.

Options (1), (3) and (4) are false

Concept:

(1) If  exists (finitely or infinitely), then 

(2) Cauchy's first theorem on limits, 

If  , then 

Explanation:

Let a_n = 

If  exists (finitely or infinitely), then 

∴  

Now by Using Cauchy's first theorem on limits,

Hence, 

Concept:

Explanation:

(A) we have

⇒ 

= 

= 27

Hence,  = 27

(B) we have

⇒ 

Hence, both the statement are true.

Option (3) is correct

Explanation:

 is convergent so  = 0

Since every convergent sequence is bounded

So {anbn} is bounded.

So at least one of {an}, {bn} is bounded  

Option (1) is correct

 an = n, bn=  so  is convergent

but ∑ an is not convergent also {an} is not bounded.

Options (2), (4) are false

Let an = bn=  

then  is convergent but both ∑ an, ∑ bn are not convergent

Option (3) is false

Concept -

(i) n³ + 1 = (n+1)(n² + 1 - n)

(ii)    and  

(iii) 

Explanation -

We have the series 

= 

=  

= 

= 

= 

= e + (e - 1) = 2e - 1

Hence the given series is convergent and cgs to finite limit 2e - 1.

Hence option(iii) is correct.

Concept:

Casaro's theorem: Let a_n and b_n be two sequence such that they converges to a and b respectively. Then

 converges to ab.

Explanation:

Let S_n =  and t_n = 

Then 

and 

Hence, by Casaro's theorem

Concept:

Explanation:

   = 

   = 

   = 

   = 

   = 

  = e - 1 - (e - 1 - 1) (as )

   = e - 1 - e + 1 + 1 = 1

Option (4) is true