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QUESTION: 1

The sequence

Solution:

QUESTION: 2

The series 1 · x + 1 · 2x^{2} + 1 · 2 3x^{2} + ... + + ... + n! · x^{n} + ... is

Solution:

**Correct Answer :- a**

**Explanation : **a_{n} = n!

then |a_{n+1}/a_{n}| = (n+1)/n!

= n+1

L = lim(n-->∞) |a_{n+1}/a^{n}|

= ∞

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 ∑μ_{n} be the given series, then we have

∴ From ratio test, the given series ∑μ_{n }is convergent or divergent according as

QUESTION: 5

If a_{1}, a_{2}, a_{3}, ... be a sequence of non-zero integers satisfying

II. a_{n} divides a_{n + 1}

III. every integer is a divisor of some a_{n}, then

Solution:

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

QUESTION: 6

Let ∑μ_{n} be a series of positive terms. Given that ∑μ_{n} 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 ∑μ_{n} be the given series, then we have

Therefore, from ratio test, we conclude that the given series ∑μ_{n} 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

∑μ_{n }whose nth term and can be proved to be divergent.

Hence, we conclude that the given series is convergent, if x^{2} ≤ 1 and divergent, if x^{2} ≥ 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.

∴ u_{n} is also convergent series.

Hence, by d’Alembert test

implies u_{n} is divergent series.

By Cauchy condensation test, ∑μ_{n} 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, v_{n} is divergent.

Hence, t_{n} is also divergent.

QUESTION: 12

The series convergent, if

Solution:

Here given series is is convergent

∴ The given series will be convergent, if

Hence, v_{n} is convergent series, so u_{n} is also convergent.

∴ The given series is convergent, if | x | ≤ 1.

QUESTION: 13

The series

Solution:

Let the given series be denoted by ∑μ_{n}, then

∑|μ_{n}|

Compare this series ∑v_{n} with the auxiliary series,

Then,

which is finite quantity.

Hence, ∑u_{n} and ∑w_{n} are either both convergent or both divergent. But ∑w_{n}

Hence, the series ∑v_{n} is divergent.

Also, in the series ∑u_{n}, 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 ∑u_{n} is convergent. H ence, the given series ∑u_{n} is conditionally convergent.

QUESTION: 14

The series

Solution:

which is finite and non-zero.

Since, ∑v_{n} is divergent, therefore, ∑u_{n} 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, ∑v_{n} is divergent and ∑u_{n} 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, ∑u_{n} is convergent.

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