Sequences And Series Of Real Numbers -4 - Mathematics MCQ

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

Option c will be correct answer of this 

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

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

So, both the series have same radius of convergence.

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.

Correct Answer :- b

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.

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.

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.

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

x_n = 2²ⁿ for all n ∈ N