Sequences And Series Of Real Numbers -4

# Sequences And Series Of Real Numbers -4

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## 20 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Sequences And Series Of Real Numbers -4

Sequences And Series Of Real Numbers -4 for IIT JAM 2023 is part of Topic-wise Tests & Solved Examples for IIT JAM Mathematics preparation. The Sequences And Series Of Real Numbers -4 questions and answers have been prepared according to the IIT JAM exam syllabus.The Sequences And Series Of Real Numbers -4 MCQs are made for IIT JAM 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Sequences And Series Of Real Numbers -4 below.
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Sequences And Series Of Real Numbers -4 - Question 1

### If then

Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 1

Explanation : For |a|<1

b = 1/a, |b|>1

lim (n→∞) an = lim (n→∞) (1/b)n

= lim(n→∞) 1/bn

= 1/(±∞)

= 0

Sequences And Series Of Real Numbers -4 - Question 2

### Which amongst the following statements is true?

Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 2

Option c will be correct answer of this . every bounded sequence is not convergent for example (-1)^n is bounded by -1 and 1 but it is not convergent .

Sequences And Series Of Real Numbers -4 - Question 3

### A Cauchy sequence is convergent, if it is a

Sequences And Series Of Real Numbers -4 - Question 4

If radius of convergence of series is 1, then radius of convergence of series is…

Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 4

By the differentiation’s theorem, we have ...(i)

Differentiating term by term, we obtain ...(ii)

So, both the series have same radius of convergence.

Sequences And Series Of Real Numbers -4 - Question 5

If {xn} and {yn} are two convergent sequence such that xn < yn, n ∈ N, then

Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 5

The correct option is Option A.

Let lim xn = x and lim yn = y and zn = yn - xn

n->∞                n->∞

Then (zn) is a convergent sequence such that zn > 0 ∇ n ∊N

and lim zn = y - x.

n->∞

Now (zn) is a convergent sequence of real numbers and zn > 0 ∇ n ∊N

So, lim zn ≥ 0

n->∞

So, y - x ≥ 0

=> y ≥ x

=> lim yn ≥ lim xn

n->∞      n->∞

Hence, proved.

Sequences And Series Of Real Numbers -4 - Question 6

Let a = min{x2 + 2x + 3, x ∈ R} and then the value of Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 6 Sequences And Series Of Real Numbers -4 - Question 7

The sequence Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 7

Explanation : Suppose otherwise, that there exists a number L implies R and a positive integer N such that

| f(n) - L | < e {for all } e > 0 {for all } n > N.

Since N is a positive integer, we know 4N > N and 4N+2 > N.

But,

f(4N) = cos (2N pi) = 1 and f(4N+2) = cos((2N+1)pi) = -1

Taking e = 1/2

= |1 - L| < 1/2 and |-1-L| < 1/2

=> |1+L| < 1/2

But, these imply

|1-L| + |1+L| < 1.

By the triangle inequality we then have

|1 - L + 1 + L | < 1

=> 2 < 1

a contradiction. Hence, there is no such limit L.

Therefore, the sequence converges to zero.

Sequences And Series Of Real Numbers -4 - Question 8

If for any e > 0, there exists a positive integer m such that I an - am | < e whenever n ≥ m . The sequence an > is called

Sequences And Series Of Real Numbers -4 - Question 9

A convergent sequence is a Cauchy sequence, if it is a

Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 9

Solution :- If {an}∞n=1 is a cauchy sequence of real numbers and if there is a sub-sequence of this sequence, {anj}∞j=1 which converges to a real number L, then I need to show that the sequence {an}∞n=1 converges to the real number L.

Sequences And Series Of Real Numbers -4 - Question 10 Sequences And Series Of Real Numbers -4 - Question 11 Sequences And Series Of Real Numbers -4 - Question 12

For the given Arithmetic progression find the position of first negative term? 50, 47, 44, 41,............

Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 12

Let nth term=0, the next term would be first negative term.
0 = 50 + (n - 1) – 3, n = 17.66.. therefore at n = 18 the first negative term would occur.

Sequences And Series Of Real Numbers -4 - Question 13

In the given AP series the term at position 11 would be? 5, 8, 11, 14, 17, 20.........50.

Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 13

nth term = a + (n – 1)d, nth term = 5 + (11 - 1)3 = 35.

Sequences And Series Of Real Numbers -4 - Question 14

The value of K for which the set of equations x + ky + 3z = 0, 3x + ky - 2z = 0, 2x + 3y - 4z = 0, has a non-trivial solution over the set of rationales is

Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 14

For non-trivial soln det of Coefficient, matrix should be 0.
|1 K 3|
|3 K 2| =  0
|2 3 4|
4K - 4K + 27 - 6K + 6 + 12K = 0
2K + 33 = 0
33 K=2
K=33/2

Sequences And Series Of Real Numbers -4 - Question 15

If sequences are convergent, then

Sequences And Series Of Real Numbers -4 - Question 16

A sequence contains a convergent subsequence, if it is

Sequences And Series Of Real Numbers -4 - Question 17 be a sequence converges to 0 and be a sequence that is bounded, then is a sequence that

Sequences And Series Of Real Numbers -4 - Question 18

Which of the following sequences of functions is uniformly convergent on (0, 1)?

Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 18 Sequences And Series Of Real Numbers -4 - Question 19

The sequence {xn}, where xn = nl/n, converge to

Sequences And Series Of Real Numbers -4 - Question 20

Let xn = 22n for all n ∈ N. Then the sequence {xn}

Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 20

xn = 22n for all n ∈ N ## Topic-wise Tests & Solved Examples for IIT JAM Mathematics

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