Sequences & Series (Convergence) MCQ Quiz in বাংলা - Objective Question with Answer for Sequences & Series (Convergence) - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

The necessary condition of convergent series  is  .

If  then the series  is divergent. 

Explanation:

(1) Let u_n = cos 

 = cos 0 = 1 ≠ 0

 is divergent.

(2) Let 

∴  = 

 = 1 ≠ 0 

Hence ∑u_n is divergent.

(3) Let 

∴  =  - 1

  - 1

 - 1

 - 1

 - 1

= ( +0 + 0 + .... + 0) - 1

= 1/3 ≠ 0

∴ ∑ u_n is divergent.

Hence option (4) is correct

Concept:

Root Test :

To apply the root test, we need to evaluate

 

If this limit is less than 1, the series converges absolutely;

if it is greater than 1, the series diverges;

and if it equals 1, the root test is inconclusive.

Explanation: 
   
Given :  
   

 Apply the root test:
   
  =   .
   
This expression can be simplified by breaking it into two parts:
   
     =  .
   

Evaluate   :

Since   grows very slowly compared to exponential terms,
   
  =   = 1.
   

Evaluate  :

Since   dominates n as , we have
   

  =   =   = 2.
   

Now,    =  
   

Since  =   

the series   converges absolutely by the root test.

Hence Option(1) is the correct answer.

Explanation: 

Let

S =  

After rearranging the first and second terms of each series, we get, 

= 

= 

Here we can see that  is geometric series with a=  and r =  

We know that the sum of infinite geometric series is  , r

Hence the sum of the given series is 

S = 

∴ The sum of the series is 3/10.

Explanation:

   = 

   = 

   = 

   = 

   = 

  = 1 - e^-1- e^-1 (as )

   =1- 2e^-1 = 1 - (2/e)

Option (2) is true

Given: 



This can be rewritten as:

 .

⇒ 

Thus, we have:

 =  > 1.

Let   is an increasing.

Let  =y and =x ,then 

y =  .

To check if y is an increasing function, we calculate  :

  =   .

For y to be increasing, we need   > 0, so

 =   >0 

⇒ 0\)

⇒ 

⇒ 

Therefore, y is a increasing function for  .

since  = 1 and   = 3 

Limit of   = [1,3].

Hence Option 3 is the correct answer.

Concept:

Dirichlet’s test: Let ∑ a_n be infinite series such that the series of partial sum S_n = a₁ + a₂ + ... + a_n is bounded. let {b_n} be a monotonically decreasing infinite sequence that converges to zero. Then ∑ a_nb_n converges to some finite value.

Limit comparasion test: Let ∑ an and ∑ bn be two series of positive term and c =  ≠ 0 and finite then ∑ an and ∑ bn converges and diverges together.

Explanation

(a): 

Here  has bounded series of partial sum 

and  is a sequence of monotonically decreasing sequence converges to 0

Then by Drichlet's test  is convergent

(a) is convergent

(b): 

Let a_n =  and b_n = 

Then  =  (0/0 form)

                      =  (L'hospital rule)

                   = 1 ≠ 0

So, by limit comparasion test, ∑ an and ∑ bn converges and diverges together.

Now, ∑ bn = ∑  is convergent series by p-series test.

Hence  is convergent

(b) is convergent

(3) is correct

Concept:

(i) Limit Comparison Test : 

If, (a) a_n ≥ 0 and b_n ≥ 0 , ∀ n

(b) , Where c is finite positive constant.

Then, the series  and  behave alike 

(ii) p- series test :

If p > 1. Then the series  converges.

If p ≤ 1. Then the series  diverges.

Explanation:

(a) Let a_n=sin() and b_n= 

Now,  0(finite) 

By using limit Comparison test

  and  behave alike 

Now, By using p - series test  diverges 

⇒  is also diverges.

(b) + = 

Let a_n=  and bn= 

​0(finite)​​

By using limit Comparison test

  and  behave alike 

Now, By using p - series test  converges. 

⇒  is also converges.

Hence, (1) option is true

Concept:

Integral test : This test is used for series of the form  where f(x)  is a continuous, positive, and decreasing function for x ≥ 1 

The series converges if and only if the corresponding improper integral  converges.

Explanation:

S_n = 

 S_n =   

= 

= log_e

= log_e2

Explanation - If every subsequence of a sequence is convergent then sequence is convergent 

Option 1:-  Let 

  and    

Here subsequence is convergent but sequence oscillate between 0 and 4 . 

Therefore , Option 1 is False.

Option 2 :- Let x_n = (-1)ⁿ then  {(-1)nxn} = {1} which is converges but {xn} does not converges

 Option 2 is false.

Option 3:- By Cauchy Second Theorem on limit

If  exist then  also exist

and so bounded. 

Option 3) is True.

Option 4 :- Let 

It is unbounded sequence but  here one subsequence is 4 which is bounded  so, Option 4 is False. 

Concept:

For |a|

Calculation:

We know that,

Substituting x by -x, we get:

Now, 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Now, 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒  [∵ - ln x = ln ()]

∴