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

The correct answer is option C.

Here's why option C is true:

• A sequence can only converge to one limit. If a sequence were to have more than one limit, it would not be converging.

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.

- Convergent sequences have limits as n→∞.
- Let sequences a_n and b_n converge to L and M respectively.

- The sequence (a_n​+b_n​) converges to L + M.

- The sequence (a_n​b_n​) converges to L x M.


- Thus, a_n + b_n is always convergent, confirming option A.

• The sequence {a_n} converges to 0, meaning that for any small positive number ε, there exists an N such that for all n > N, |a_n| < ε.


• The sequence {b_n} is bounded, indicating there is a positive number M such that for all n, |b_n| < M.


• The product sequence {a_nb_n} is given by a_nb_n = a_n * b_n.


• Since |a_n| < ε and |b_n| < M, we have |a_nb_n| < εM.


• As n approaches infinity, |a_nb_n| approaches 0, showing that {a_nb_n} converges to 0.


 

x_n = 2²ⁿ for all n ∈ N