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QUESTION: 1

The sequence <s_{n}> = is

Solution:

Here lower bound = l = − 1/2 , & upper bounded u = 1. But the given sequence is neither increasing nor decreasing. Thus, <s_{n}> is bounded but not monotonic.

QUESTION: 2

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

Solution:

The sequence of n th partial sum of given series is unbounded and goes to infinity.

since sequence of n th partial sum is unbounded, it is not convergent therefore the given series

is not convergent i.e. divergent.

QUESTION: 3

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: 4

The arithmetic mean between 2 + √(2) and 2 - √(2) is

Solution:

QUESTION: 5

For the sequence 1, 7, 25, 79, 241, 727 … simple formula for {a_{n}} is ____________

Solution:

The ratio of consecutive numbers is close to 3. Comparing these terms with the sequence of {3^{n}} which is 3, 9, 27 …. Comparing these terms with the corresponding terms of sequence {3^{n}} and the nth term is 2 less than the corresponding power of 3.

QUESTION: 6

The interval of convergence of is

Solution:

R = 1 / P = 1 = radius of convergence

When x = 1 & x = –1, Then is converges at both

x = 1 & x = – 1.

so, its interval of convergence is exactly [–1, 1].

QUESTION: 7

The sequence

Solution:

We have

QUESTION: 8

The sequence

Solution:

QUESTION: 9

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: 10

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: 11

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: 12

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: 13

The series

Solution:

which is finite and non-zero.

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

QUESTION: 14

The series is

Solution:

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

QUESTION: 15

Solution:

Here, it is given that

Hence, ∑v_{n} is divergent and ∑u_{n} is also divergent and

QUESTION: 16

The sequence {S_{n}} of real numbers given by S_{n }= is

Solution:

So <S_{n}> is monotonically increasing Next, we will show that it is bounded.

⇒ <S_{n}> is bounded. By the theorem's Every monotonic bounded sequence is convergent, Then <S_{n}> is convergent.

By the Lemma, if <S_{n}> is a convergent sequence of real numbers, Then <S_{n}> is a Cauchy-sequence.

QUESTION: 17

The series

Solution:

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

QUESTION: 18

The sequence 1, 1, 1, 1, 1…. is?

Solution:

For limit n tending to infinity the sum also tends to infinity and thus it is not summable.

QUESTION: 19

The series

Solution:

The nth term of the given series is

Hence, ∑u_{n} is convergent.

QUESTION: 20

Solution:

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