Sequences And Series Of Real Numbers -10 - Mathematics MCQ

A sequence is an ordered list of numbers, and each number in the sequence is called an element or a term of the sequence. The set of all distinct elements in a sequence is called the range set of the sequence.

The given series can be written as where a_3k = 2^–k & a_n = 0 if n is not multiple of 3.


Therefore the radius of convergence is

Correct Answer :- B

Explanation : For each n ∈ N, apply AM-GM inequality for a₁ = 1, a₂ = a₃ = .... =a_n+1

= 1 + 1/n.

 We get e_n+1 > e_n is increasing and bounded.

The difference in any term with the previous term is same.

The sequence x_n​ is defined as the nth harmonic number, which is the sum of the reciprocals of the positive integers up to n:

Let's consider each statement:

I. A Cauchy sequence is a sequence where for every positive real number ε, there is an integer N such that for all m,n>N, the absolute difference ∣x_n​−x_m​∣ is less than ε. For the harmonic sequence, the difference between terms does not eventually become arbitrarily small because as n grows larger, the terms being added to the sum 1/n get smaller, but there's an infinite number of them, so the sum continues to grow without bound. Therefore, x_n​ is not a Cauchy sequence.

II. The harmonic series is divergent, which means that as n approaches infinity, x_n​ increases without bound and does not converge to a limit. Therefore, the sequence x_n​ is not convergent.

III. A sequence is bounded if there is a real number M such that for all n, ∣x_n​∣ ≤ M. The harmonic sequence is not bounded because it increases without limit as n approaches infinity.

Given these points, the correct statements are:

II. x_n​ is not convergent. III. x_n​ is not bounded.

The sequence x_n​ is indeed not a Cauchy sequence, but the statement is not given as an option, so we do not consider it in the multiple-choice answers.

Here f : R → R is strictly increasing continuous function. If, {a_n} is a sequence in [0, 1]. 
Then {a_n} is a bounded sequence. 
f is continuous on R, Then {f(a_n)} is a bounded sequence by properties of continuous functions. 
Next {a_n} is a sequence in [0, 1], So it is not necessary that {a_n} is convergent

Here, s₁ = 1 < a + 1

 1 ≤ s₁ < a + 1

ψ  1 ≤ s₂ = < a + 1

Let 1 ≤ s₁ < a + 1, Then

1 ≤  s_m+1

By the mathematical induction,

                   1 ≤ s_n < a + 1

⇒ <s_n> is bounded.

we can also show that <s_n> is monotonically increasing.

Therefore the sequence <s_n> is convergent.

Let lim s_n = 1, Then

⇒ l =

⇒ l² – l – a = 0
 

The above equation has only one positive root ≥ 1.

Hence, the sequence <s_n> converges to the positive

roots of x² – x – a = 0.