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I am given that $\sum\limits_{n=1}^\infty a_n$ is convergent.

I need to determine whether $\sum\limits_{n=1}^\infty (a_n)^\frac{1}{3}\;$ and $\;\sum\limits_{n=1}^\infty (a_n)^2\;$ are also convergent.

Imagine that $a_n = \dfrac{1}{n^4}.\;$ I believe that this is convergent because it's converging to $0$.

Following the same thought, if $\displaystyle a_n = \left(\frac{1}{n^4}\right)^2,\;$ it's convergent because it's converging to $0$.

Am I doing this correctly or there is some other way to prove this?

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That's not true for, e.g. for $a_n=n^{-2}$. Then $\sum a_n<\infty$ but $\sum a_n^{1/3} = \infty$. Another example is $a_n = -n^{-1/2}$, then $\sum a_n<\infty$ but $\sum a_n^2 = \infty$. –  Ilya Jan 3 '13 at 12:52
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@Ilya: Your second example should be $a_n=\frac{(-1)^n}{\sqrt{n}}$. –  Mhenni Benghorbal Jan 3 '13 at 13:21
    
@MhenniBenghorbal: thanks, you're right. –  Ilya Jan 3 '13 at 13:42
    
@Ilya: You are welcome. –  Mhenni Benghorbal Jan 3 '13 at 13:59

3 Answers 3

up vote 5 down vote accepted

Task: You need to establish whether the statements are true for all $(a_n)$.

  • To prove such a statement is true, you need to show it is always holds for any convergent $a_n$ (not just that it holds for some $a_n$).
  • But to conclude that a given statement is false, you can simply find a single $a_n$ that serves as a counterexample.

Clarification: Recall that IF a series $\sum a_n$ converges, THEN $\lim a_n \to 0.\quad(1)$

The converse of (1) does not hold: if $\lim_{n\to \infty} b_n = 0$, it doesn't necessarily follow that $b_n$ converges.


You can use the p-series test to find an $a_n$ such that$\sum_{n=1}^\infty a_n$ converges, but such that $\sum_{n=1}^\infty (a_n)^{1/3}$ diverges.

p-series test: Recall that for $a_n = \dfrac{1}{n^p}, \;$ $\displaystyle \sum_{n=1}^\infty \frac{1}{n^p}\;$ converges if $p > 1$, and diverges if $p \le 1$.

E.g. $p = 3$, $a_n = \dfrac{1}{n^3}\implies \sum_{n=1}^\infty \dfrac{1}{n^3}$ converges.

Then $(a_n)^{1/3}$ gives $\sum_{n=1}^\infty \left(\dfrac{1}{n^3}\right)^{1/3} =\;\; \sum_{n=1}^\infty \dfrac{1}{n},\;\;$ which diverges.


Now we need to check whether the fact that $\sum_{n=1}^{\infty}a_n$ converges, implies that $\sum_{n=1}^{\infty}a_n^2$ converges. This is more of a challenge, since it seems to follow that, yes, it must.

Here, we have to get creative to find a counterexample, if one exists:

If we choose $\sum a_n$ to be an alternating series (sign of terms change depending on even, odd $n$) and so that $\sum_{n=1}^{\infty}a_n$ would converge but not $\sum_{n=1}^{\infty}a_n^2$, we are in luck.

Let $ a_n=\dfrac{(-1)^n}{\sqrt{n}}$ (the sum of which converges, non-absolutely, since some terms are positive, others negative).

Then $(a_n)^2 = \left(\dfrac{(-1)^n}{n^{1/2}}\right)^2 = \dfrac{1}{n}$; this sum, with all positive terms, you will recognize to be the harmonic series, which you know diverges.

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The $p$-series approach is I think the easiest to work with here. Let $a_n = \dfrac{1}{n^p}$. Then the OP wants to find a value of $p$ such that $p > 1$, but $\frac{p}{3} \leq 1$. Nice post (as usual!). –  JavaMan Jan 3 '13 at 16:15
    
@amWhy Sorry for the late feedback. Thanks for this amazing explanation. The examples provided where essential for my understanding. Many many thanks –  Favolas Jan 3 '13 at 20:43
    
You're very welcome, Favolas! –  amWhy Jan 3 '13 at 20:45

Youu want to decide whether or not the statements are true for any sequence $(a_n)$. So you either prove them, or provide a counterexample.

What you have done is wrong on many counts. You took a sequence $a_n$ and said (I believe) $\sum_{n=1}^{\infty}a_n$ converges because $a_n\to 0$. This is wrong as the harmonic series $\sum_{n=1}^{\infty}\frac1n$ diverges although $\frac1n\to 0$. 

For an easier counterexample take $a_n=\frac{1}{n^3}$ for the first problem. Indeed, $\sum_{n=1}^{\infty}\frac1{n^3}$ converges as  $$0\le \frac1{n^3}\le \frac1{n^2}$$ and the second series converges (why?). You can also use the integral test or the Cauchy Condensation test. What is $$\sum_{n=1}^{\infty}\left(\frac1{n^3}\right)^{\frac13}?$$

The second problem is trickier: You want $\sum_{n=1}^{\infty}a_n$ to converge but not $\sum_{n=1}^{\infty}a^2_n$. This creates a small problem:

Because $a_n\to 0$, for large $n$, $\left|a_n\right|\ge a^2_n$ (why?) which should mean that the  second series would converge. Not so much, if we choose $a_n$ to alternate signs. Then  $\sum_{n=1}^{\infty}a_n$ would converge but not $\sum_{n=1}^{\infty}a^2_n$ as in the second series the terms would only be added, while in the first series  terms are also subtracted (alternating signs). An example is $a_n=\frac{(-1)^n}{\sqrt{n}}$

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Hi Nameless. Once again, thanks for your explanation. –  Favolas Jan 3 '13 at 20:37

$\sum\limits_{n=1}^\infty\frac{1}{n^3}$ & $\sum\limits_{n=1}^\infty(-1)^n\frac{1}{\sqrt n}$ are convergent but $\sum\limits_{n=1}^\infty\frac{1}{n}$ is not.

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