## A Bonus Question on Convergent Series

Occasionally when teaching the sequences and series material in second-semester calculus I’ve included the following question as a bonus:

Question: Suppose $\displaystyle \sum_{n=1}^{\infty} a_n$ is absolutely convergent.  Does that imply anything about the convergence of $\displaystyle \sum_{n=1}^{\infty} a^2_n$?

The answer is that $\displaystyle \sum_{n=1}^{\infty}a^2_n$ converges.  I’m going to give two proofs.  The first is the more straightforward approach, but it is somewhat longer.  The second is clever and shorter.

First method: If $\displaystyle \sum_{n=1}^{\infty} a_n$ is absolutely convergent, then, by the divergence test, $\displaystyle \lim_{n \to \infty} |a_n| = 0$.  Thus there exists some $N > 0$ such that if $n \geq N$ then $|a_n| \leq 1$.  This means that, for $n \geq N$, $a^2_n \leq |a_n|$.  By the direct comparison test, then, $\displaystyle \sum_{n=1}^{\infty} a^2_n$ converges.

Second method: As we argued above, $\displaystyle \lim_{n \to \infty} |a_n| = 0$.  Thus $\displaystyle \lim_{n \to \infty} \frac{a^2_n}{|a_n|} = \lim_{n \to \infty} |a_n| = 0$.  By the limit comparison test, then, $\sum_{n=1}^{\infty} a^2_n$ converges.

(The first method is David Mitra’s answer and the second method is my answer to this Math.SE question.)

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### 6 Responses to A Bonus Question on Convergent Series

1. velikod says:

I admire your work, you pick very nice problems! I just want to ask something. How are you writing formulas? Please don’t tell me that you generate them as pics and insert in the code.. I think there is WP plugin that works with MathJax so that you write code in latex and it is rendered nicely.

• mzspivey says:

I use the LaTeX environment in WordPress.

• velikod says:

I am sorry then. You can delete my comments if you wish.

• mzspivey says:

No need to apologize. And thanks for your interest in my work!

2. pluprofedgar says:

How about this (assuming the terms of the series positive so I can avoid absolute value signs): For all $n\in \mathbb{N}$, $a_1^2+a_2^2+\cdots +a_{n-1}^2+a_n^2\leq (a_1+a_2+\cdots +a_{n-1}+a_n)^2$. Thus $\lim_{n\to\infty}\sum_{i=1}^na_i^2 \leq \lim_{n\to\infty}(\sum_{i=1}^na_i)^2 = L^2$ where $L=\sum_{i=1}^\infty a_i$. Thus the sequence of partial sums of $(a_1^2,a_2^2,\ldots)$ is increasing and bounded above by $L^2$ so the series converges.