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A Convergent series of real numbers $\sum a_{n}$ is given , what can be said about the convergence of $\sum a_{n}^{2}$ and $\sum a_{n}^{4}$.

Also , if only absolute convergence of $\sum a_{n}$ , i.e. convergence of $\sum |a_{n}|$ is given what about convergence of $\sum a_{n}^{2}$ and $\sum a_{n}^{4}$.

Please give me some hints as to how to proceed.

Thanks.

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    $\begingroup$ I would say, look at the definition of convergence $\endgroup$ – 9301293 Sep 6 '15 at 20:53
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    $\begingroup$ "If only absolute convergence ... is given " doesn't make sense: absolute convergence is a stronger property than convergence. If you have absolute convergence of $\sum a_n$, then convergence of $\sum a_n^m$ follows from the comparison test for any $m \ge 1$. $\endgroup$ – Rob Arthan Sep 6 '15 at 21:07
  • $\begingroup$ @RobArthan : Yes , that's right $\endgroup$ – user118494 Sep 6 '15 at 21:24
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    $\begingroup$ So why not edit your question to remove the confusing word "only"? $\endgroup$ – Rob Arthan Sep 6 '15 at 21:26
  • $\begingroup$ @RobArthan : Why remove that $?$ Whoever is going to answer or commenters like you definitely know this particular topic better than me . I prefer to let them know all my silly confusions. $\endgroup$ – user118494 Sep 6 '15 at 21:39
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If all that you know is that $\sum a_n$ converges, nothing can be said about the convergence of $\sum a_n^2$ and $\sum a_n^4$.

Proexample: Let $a_n=0$. Here $\sum a_n$ converges, and $\sum a_n^2$ and $\sum a_n^4$ converge.

Counterexample: Let $a_n=(-1)^n\frac{1}{\sqrt[4]{n}}$. Here $\sum a_n$ converges by the alternating series test, but neither $\sum a_n^2$ nor $\sum a_n^4$ converge ($p$-test).

If, however, you know that $\sum a_n$ converges absolutely, then $\sum a_n^2$ and $\sum a_n^4$ must converge absolutely as well, so a fortiori they must converge.

Proof: Since the series $\sum a_n$ converges, the sequence $a_n$ must also converge. Thus there must exist an $N\geq 0$ such that $n \geq N$ implies $|a_n|<1$. Since $\sum |a_n|$ converges, we can define $S=\sum_{n=N}^\infty |a_n|$. Then, as $|a_n^2|\leq|a_n|$ and $|a_n^4|\leq|a_n|$ for $n\geq N$, we have $\sum_{n=0}^\infty |a_n^2|\leq\sum_{n=0}^{N-1}|a_n^2|+S$ and $\sum_{n=0}^\infty |a_n^4|\leq\sum_{n=0}^{N-1}|a_n^4|+S$, both of which are finite. Thus $\sum a_n^2$ and $\sum a_n^4$ converge absolutely, which implies that they converge.

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Hint:

Look at the two cases of $a_n=\frac{(-1)^n}{n^{1/4}}$ and $a_n=\frac{1}{n^2}$ for the first question.

Now, for absolute convergence (or equivalently if $a_n$ is non-negative), use the sandwich theorem to show convergence of the (non-negative) series of general term $a_n^2$ and $a_n^4$. (For $n$ big enough, $\lvert a_n\rvert < 1$: this will be useful).

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If $\sum a_n$ is convergent, then is $\sum a_n^2$, $\sum a_n^4$, or, for the matter of fact, $\sum a_n^k$ for any $k > 1$, convergent too. To see this, since $\sum a_n$ is convergent, then $|a_n| < 1$ for all $n > N$. Then $|a_n^k| < |a_n|$ if $k > 1$ and $$\sum a_n^k < \sum a_n,$$ thus $\sum a_n^k$ is convergent.

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  • $\begingroup$ Are you assuming $\sum a_n$ is absolutely convergent? $\endgroup$ – zhw. Dec 14 '15 at 2:33
  • $\begingroup$ @zhw Yes, why..? $\endgroup$ – user98186 Dec 14 '15 at 14:43
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    $\begingroup$ You said "if $\sum a_n$ is convergent", which is not the same thing. $\endgroup$ – zhw. Dec 14 '15 at 17:52
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I know one result which says that product of two convergent series is convergent if atleast one of the two series converges absolutely.Use this result.

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Consider the following property of a converging series:$$\lim_{n\to\infty}a_n=0$$Because if a series does converge, then $$\lim_{m\to\infty}\sum_{n=0}^{m}a_n=\lim_{m\to\infty}\sum_{n=0}^{m+1}a_n$$Due to the properties of $\infty$. We then get $$\lim_{m\to\infty}a_{m+1}=0$$which is equivalent to my above statement.

Now consider $\sum a_n^2$. $\lim_{n\to\infty}a_n^2=0$ In particular, it will approach $0$ faster than $a_n$. So while $\sum a_n^2$ may produce values that are larger than $\sum a_n$, eventually, both summations will converge on some value.

If the first sum converges, then its squares will converge faster because $\lim_{n\to\infty}\frac{a_n^2}{a_n}=\lim_{n\to\infty}a_n=0$, meaning $a_n^2$ approaches $0$ faster.

However, the squares may not approach the same value as the original sum.

Also consider a hyper-cube/prism with n-dimensions. You want to measure the longest diagonal that goes from one vertex to the opposite vertex. Given side lengths $a_1,a_2,a_3,a_4,\cdots$, the solution would be $S=\sqrt{\sum a_n^2}$ where $S$ is the length of the longest diagonal.

In the same hypercube, we note $\sum a_n>S$.

This is all just manipulations of Pythagorean theorem for multiple side lengths along with the theory on triangles stating $$a+b>c$$where c is the longest side length.

I don't know anywhere $\sum a_n^4$ would be applied.

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