Sequences And Series Of Real Numbers -1


20 Questions MCQ Test IIT JAM Mathematics | Sequences And Series Of Real Numbers -1


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This mock test of Sequences And Series Of Real Numbers -1 for IIT JAM helps you for every IIT JAM entrance exam. This contains 20 Multiple Choice Questions for IIT JAM Sequences And Series Of Real Numbers -1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Sequences And Series Of Real Numbers -1 quiz give you a good mix of easy questions and tough questions. IIT JAM students definitely take this Sequences And Series Of Real Numbers -1 exercise for a better result in the exam. You can find other Sequences And Series Of Real Numbers -1 extra questions, long questions & short questions for IIT JAM on EduRev as well by searching above.
QUESTION: 1

The sequence <sn> =  is 

Solution:

Here lower bound = l = − 1/2 , & upper bounded u = 1. But the given sequence is neither increasing nor decreasing. Thus, <sn> 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 ∑μn be the given series, then we have


∴ From ratio test, the given series ∑μis 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 {an} is ____________ 

Solution:

The ratio of consecutive numbers is close to 3. Comparing these terms with the sequence of {3n} which is 3, 9, 27 …. Comparing these terms with the corresponding terms of sequence {3n} 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.
∴ 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: 10

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: 11

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: 12

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: 13

The series

Solution:




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


QUESTION: 16

The sequence {Sn} of real numbers given by Sn = is

Solution:


So <Sn> is monotonically increasing Next, we will show that it is bounded.

⇒ <Sn> is bounded. By the theorem's Every monotonic bounded sequence is convergent, Then <Sn> is convergent.
By the Lemma, if <Sn> is a convergent sequence of real numbers, Then <Sn> 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, ∑un is convergent.

QUESTION: 20

Solution: