Sequences And Series Of Real Numbers -1


20 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Sequences And Series Of Real Numbers -1


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QUESTION: 1

​The sequence 

Solution:


QUESTION: 2

The series 1 · x + 1 · 2x2 + 1 · 2 3x2 + ... + + ... + n! · xn + ... is

Solution:

Correct Answer :- a

Explanation : an = n!

then |an+1/an| = (n+1)/n!

= n+1

L = lim(n-->∞) |an+1/an|

= ∞

The radius of convergence R = 0 Hence, the series converges only at x = 0.

Option A) is correct, it is divergent.

QUESTION: 3

The series 2 + 4 + 6 + 8 + ...is

Solution:

QUESTION: 4

The scries x  is convergent, if

Solution:

If ∑μn be the given series, then we have


∴ From ratio test, the given series ∑μis convergent or divergent according as 

QUESTION: 5

If a1, a2,  a3,  ... be a sequence of non-zero integers satisfying

 
II. an divides an + 1  
III. every integer is a divisor of some an, then 

Solution:




 is a contradiction chair, and so, A is an irrational number.

QUESTION: 6

Let ∑μn be a series of positive terms. Given that ∑μn is convergent and also  exists, then the said limit is

Solution:


unique quantity.
Then, from Eq. (i) we have 

QUESTION: 7

The Series 

Solution:

Here, given series is

∴ If ∑μn be the given series, then we have


Therefore, from ratio test, we conclude that the given series ∑μn is convergent or divergent according as

i. e., according as x < 1 or > 1 ...(i)
If n = 1, this test fails and the given series reduces to 
∑μwhose nth term   and can be proved to be divergent.
Hence, we conclude that the given series is convergent, if x2 ≤ 1 and divergent, if x2 ≥ 1.

QUESTION: 8

The sequence 

Solution:

We have 

QUESTION: 9

The sequence 

Solution:

QUESTION: 10

Match list I with list II and select the correct answer

Solution:



is also convergent series.
∴ un is also convergent series.

Hence, by d’Alembert test 

implies un is divergent series.

By Cauchy condensation test, ∑μn is convergent series.
(D) 
By Leibnitz’s test, the series is convergent.
Also,  is divergent series. 
So, the given series is converges conditionally.

QUESTION: 11

The series whose nth term is 

Solution:

Here it is given that

Now, taking auxiliary series 

Which is finite and non-zero. Since, vn is divergent.
Hence, tn is also divergent.

QUESTION: 12

The series  convergent, if

Solution:

Here given series is  is convergent

∴ The given series will be convergent, if



Hence, vn is convergent series, so un is also convergent.
∴ The given series is convergent, if | x | ≤ 1.

QUESTION: 13

The series 

Solution:

Let the given series be denoted by ∑μn, then
∑|μn|

Compare this series ∑vn with the auxiliary series,

Then,

which is finite quantity.
Hence, ∑un and ∑wn are either both convergent or both divergent. But ∑wn 

Hence, the series ∑vn is divergent.
Also, in the series ∑un, we find that its term are alternately positive and negative, its terms are continuously decreasing and

Thus, all condition of Leibnitz’s test are satisfied and as such ∑un is convergent. H ence, the given series ∑un is conditionally convergent.

QUESTION: 14

The series

Solution:




which is finite and non-zero.
Since, ∑vn is divergent, therefore, ∑un is also divergent.

QUESTION: 15

The series 

Solution:


Hence, by Cauchy’s root test, the series convergent.

QUESTION: 16

Solution:

Here, it is given that 


Hence, ∑vn is divergent and ∑un is also divergent and 


QUESTION: 17

The series 

Solution:

Here, 
Let 

QUESTION: 18

The series 

Solution:



Hence, by d’Alembert’s ratio test, it is convergent.

QUESTION: 19

The series  convergent, if

Solution:


QUESTION: 20

The series 

Solution:

The nth term of the given series is

Hence, ∑un is convergent.