Question

Let \( \sum_{n \geq 1} a_n \) be a convergent series of positive real numbers. Then which of the following statement(s) is (are) true?

• A. \( \sum_{n \geq 1} (a_n)^2 \) is always convergent
• B. \( \sum_{n \geq 1} \sqrt{a_n} \) is always convergent
• C. \( \sum_{n \geq 1} \frac{\sqrt{a_n}}{n} \) is always convergent
• D. \( \sum_{n \geq 1} \frac{\sqrt{a_n}}{n^{1/4}} \) is always convergent

Hint:

For a convergent series, examine the growth of the terms to determine if derived series, such as squares or square roots, will also converge.

Answer 1

The Correct Option is A, C

Given: $\sum_{n\geq 1} a_n$ is a convergent series of positive real numbers.

This means $a_n \to 0$ as $n \to \infty$.

Option (A): $\sum_{n\geq 1} (a_n)^2$ is always convergent

Counterexample: Let $a_n = \frac{1}{n \ln n}$ for $n \geq 2$ (and $a_1 = 1$).

The series $\sum_{n\geq 2} \frac{1}{n \ln n}$ diverges (integral test), so this doesn't work.

Let me try $a_n = \frac{1}{n^{3/2}}$. Then $\sum a_n$ converges (p-series with $p = 3/2 > 1$).

Now $(a_n)^2 = \frac{1}{n^3}$, and $\sum \frac{1}{n^3}$ converges.

But we need to check if this is always true. Since $\sum a_n$ converges, we have $a_n \to 0$.

For large $n$, if $a_n < 1$, then $(a_n)^2 < a_n$.

By the comparison test, since $(a_n)^2 \leq a_n$ for sufficiently large $n$ and $\sum a_n$ converges, we have $\sum (a_n)^2$ converges.

Option (A) is TRUE 

Option (B): $\sum_{n\geq 1} \sqrt{a_n}$ is always convergent

Counterexample: Let $a_n = \frac{1}{n^2}$. Then $\sum a_n = \sum \frac{1}{n^2}$ converges.

But $\sqrt{a_n} = \frac{1}{n}$, and $\sum \frac{1}{n}$ diverges (harmonic series).

Option (B) is FALSE 

Option (C): $\sum_{n\geq 1} \frac{\sqrt{a_n}}{n}$ is always convergent

Since $\sum a_n$ converges with positive terms, $a_n \to 0$.

For large $n$, we can write $a_n < 1$, so $\sqrt{a_n} < 1$.

However, we need a stronger result. By Cauchy-Schwarz inequality: $$\left(\sum_{n=1}^N \frac{\sqrt{a_n}}{n}\right)^2 \leq \left(\sum_{n=1}^N a_n\right) \left(\sum_{n=1}^N \frac{1}{n^2}\right)$$

Since both $\sum a_n$ and $\sum \frac{1}{n^2}$ converge, the right side is bounded.

Therefore, $\sum_{n=1}^N \frac{\sqrt{a_n}}{n}$ is bounded, and since terms are positive, the series converges.

Option (C) is TRUE 

Option (D): $\sum_{n\geq 1} \frac{\sqrt{a_n}}{n^{1/4}}$ is always convergent

Counterexample: Let $a_n = \frac{1}{n^{3/2}}$. Then $\sum a_n$ converges (p-series with $p = 3/2 > 1$).

Now: $$\frac{\sqrt{a_n}}{n^{1/4}} = \frac{n^{-3/4}}{n^{1/4}} = \frac{1}{n}$$

And $\sum \frac{1}{n}$ diverges.

Option (D) is FALSE 

Answer: (A) and (C)