Series

**nmatthies1** · May 31st 2009, 06:05 PM

Is it true that $\sum_{n=1}^{\infty} {a_{n}}^2$ is convergent if $\sum_{n=1}^{\infty} {a_{n}}$ is convergent? Justify your answer.

Any hints?

**Jose27** · May 31st 2009, 06:13 PM

Remember that if $\vert a \vert \leq 1$ then $a^2 \leq \vert a \vert$ . Now also remember if the series is convergent what is $\lim_{n \to \infty} \ a_n$ ?

Although now that I think of it, I'm not sure it works in general if the series does not converge absolutely...if it does, following the above idea suffices, but if the series converges only conditionally I'm not so sure.

**Sampras** · May 31st 2009, 06:18 PM

Originally Posted by Jose27

Remember that if $a \leq 1$ then $a^2 \leq a$ . Now also remember if the series is convergent what is $\lim_{n \to \infty} \ a_n$ ?

Although now that I think of it, I'm not sure it works in general if the series does not converge absolutely...if it does, following the above idea suffices, but if the series converges only conditionally I'm not so sure.

no if $a = -8$ , then $a^2 = 64 \not \leq -8$ ....

**putnam120** · May 31st 2009, 06:19 PM

Yes, by the comparison test.
Since $\lim_{n\to\infty}a_n=0$ then there exists $N$ such that $|a_n|<1\Longrightarrow a_n^2<|a_n|$ for all $n>N$ . Then apply the comparison test, but you can assume that $a_n\ge 0$ .

EDIT:** Sorry but the answer is NO. Take $a_n=\frac{(-1)^n}{\sqrt{n}}$ **

**nmatthies1** · May 31st 2009, 06:20 PM

Well, if the series is convergent then $\lim_{n \to \infty} \ a_n = 0$ . But how do I use this to answer the question?

**nmatthies1** · May 31st 2009, 06:23 PM

oh ok comparison test. thanks!

**Jose27** · May 31st 2009, 06:23 PM

Originally Posted by putnam120

Yes, by the comparison test.
Since $\lim_{n\to\infty}a_n=0$ then there exists $N$ such that $|a_n|<1\Longrightarrow a_n^2<|a_n|$ for all $n>N$ .

So $\sum_{n=N}^\infty a_n^2<\sum_{n=N}^\infty a_n<\infty$ .

But there's a problem, if for instance $\{a_n \}$ alternates sign, comparison test no longer applies since $\{a_n^2 \}$ is always positive, but $\{a_n \}$ isn't. Here's an example if $\{a_n \}= \{\frac{(-1)^n}{n} \}$ then $1-\frac{1}{2} \ngeq 1+\frac{1}{4}$

**NonCommAlg** · May 31st 2009, 10:26 PM

Originally Posted by nmatthies1

Is it true that $\sum_{n=1}^{\infty} {a_{n}}^2$ is convergent if $\sum_{n=1}^{\infty} {a_{n}}$ is convergent? Justify your answer.

Any hints?

it's false. counter-example: $a_n=\frac{(-1)^n}{n^c},$ where $c$ is any real number in the interval $(0, 1/2].$

**Deadstar** · June 1st 2009, 03:43 AM

Just thought i'd add in this one point...

If $\sum_n a_n$ is a convergent series of POSITIVE TERMS then $\sum_n a_n^2$ converges...

Proof,

$a_n$ must be bounded, i.e. for some $M > 0$ we have an $a_n \leq M$ for all n. Then $a_n^2 \leq Ma_n$ so by comparison test $a_n^2$ converges.

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