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.

  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.

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