Sequences And Series Of Real Numbers -4

QUESTION: 1

If then

A.
| a | ≥ 1
B.
| a | ≤ 1
C.
| a | > 1
D.
| a | < 1

Solution:

Correct Answer :- D

Explanation : For |a|<1

b = 1/a, |b|>1

lim (n→∞) a_n = lim (n→∞) (1/b)ⁿ

= lim(n→∞) 1/bⁿ

= 1/(±∞)

= 0

QUESTION: 2

Which amongst the following statements is not true?

A.
A sequence cannot converge to more than one limit
B.
Every convergent sequence is bounded
C.
Every bounded sequence is convergent
D.
Limit of a sequence is unique

Solution:

QUESTION: 3

A Cauchy sequence is convergent, if it is a

A.
sequence of real number
B.
sequence of rational numbers
C.
sequence o f irrational numbers
D.
bounded sequence of rational numbers

Solution:

QUESTION: 4

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

A.
2
B.
3
C.
1
D.
0

Solution:

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

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

So, both the series have same radius of convergence.

QUESTION: 5

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

A.
B.
C.
D.
None of these

Solution:

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.

QUESTION: 6

Let a = min{x² + 2x + 3, x ∈ R} and then the value of

A.
B.
C.
D.
None of the above

Solution:

QUESTION: 7

The sequence

A.
divergent
B.
convergent to 0
C.
convergent to 1
D.
None of these

Solution:

Correct Answer :- a

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.

QUESTION: 8

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

A.
monotonic sequence
B.
convergent sequence
C.
cauchy sequence
D.
divergent sequence

Solution:

QUESTION: 9

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

A.
sequence of rational numbers
B.
sequence of real numbers
C.
sequence of irrational numbers
D.
None of the above

Solution:

ANSWER :- b

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.

QUESTION: 11

A.
is not bounded
B.
may or may not be bounded
C.
convergent
D.
may or may not be convergent

Solution:

QUESTION: 12

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

A.
17
B.
18
C.
20
D.
None of the mentioned

Solution:

Let n^th 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.

QUESTION: 13

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