Home » Blog » If ∑ |an| converges prove ∑ an2 converges

If ∑ |an| converges prove ∑ an2 converges

Prove that if the series \sum |a_n| converges then the series \sum a_n^2 also converges. Also, give an example to show that the converse is false, i.e., a case in which \sum a_n^2 converges but \sum |a_n| does not.


Proof. Assume \sum |a_n| converges. Then we know \lim_{n \to \infty} |a_n| = 0. Thus, by the definition of the limit, for all \varepsilon > 0 there exists an integer N such that

    \[ |a_n| < \varepsilon \qquad \text{whenever} \qquad n \geq N. \]

Taking \varepsilon = 1 we then know there exists an N such that

    \[ |a_n| < 1 \qquad \text{whenever} \qquad n \geq N. \]

But then, a_n^2 = |a_n|^2 < |a_n| if n \geq N. Thus, we have

    \[ \sum_{n=1}^{\infty} a_n^2 = \sum_{n=1}^N a_n^2 + \sum_{n=N+1}^{\infty} a_n^2. \]

The first sum is a finite sum so it converges, and the second sum converges by comparison with \sum |a_n|. Hence, \sum a_n^2 converges. \qquad \blacksquare

Counterexample. The converse is false. Let a_n = \frac{1}{n}. Then,

    \[ \sum_{n=1}^{\infty} a_n^2 = \sum_{n=1}^{\infty} \frac{1}{n^2} \]

converges. However,

    \[ \sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{\infty} \left| \frac{1}{n} \right| = \sum_{n=1}^{\infty} \frac{1}{n} \]

diverges.

One comment

  1. M. Furquan says:

    Alternative proof:
    \sum_{n=1}^{N}a_n^2\le\left(\sum_{n=1}^N|a_n|\right)^2\le\left(\sum_{n=1}^\infty|a_n|\right)^2
    hence, the convergence of \sum_{n=1}^\infty|a_n| implies that the sequence of partial sums of the series \sum_{n=1}^{N}a_n^2 is bounded from above. By Theorem 10.7, the series converges.

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):