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The question is broken down into:

(i) Prove the statement

(ii) Hypothesis that $a_n \geq 0$ for each $n$ is necessary. Show that for each $p \geq 1$, $\exists (a_n)$ s.t. $\Sigma a_n$ converges but $\Sigma |a_n^p|$ diverges.

(iii) Hypothesis that $p \geq 1$ is necessary. Show that $ \forall p < 1$, $\exists$ sequence $a_n$ of nonnegative real numbers such that $\Sigma a_n$ converges but $\Sigma a_n^p$ diverges.

I'm completely stumped at how to solve all parts of this question. I'm not sure if (ii) - (iii) were supposed to help with (i). I know that from a theorem that $\ (c_n)^a$, $\Sigma c_n( x-a)^n$ converges if $|x-a|$ < $ \lim \sup |c_n|^\frac{-1}{n}$.

But I don't see how if $\Sigma a_n$ converges then so does $ \Sigma a^p_n$. Guidance on how to approach this would be much appreciated!

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  • $\begingroup$ Do you know limit comparison test?? $\endgroup$ – tattwamasi amrutam Nov 9 '16 at 0:43
  • $\begingroup$ @tattwamasiamrutam My definition of comparison test would be that if $ \Sigma a_n$ converges absolutely, and $\exists N_0$ such that $|b_n| < |a_n| \forall N,$ then $ \Sigma b_n$ converges. $\endgroup$ – Nikitau Nov 9 '16 at 0:46
  • $\begingroup$ I am saying about this: en.wikipedia.org/wiki/Limit_comparison_test. Read the one side limit test . $\endgroup$ – tattwamasi amrutam Nov 9 '16 at 0:47
  • $\begingroup$ @tattwamasiamrutam Oh sorry, yes. In my book it's listed as if $|a_n| \leq c_n$, for $n \geq N_0$ where $N_0$ is a fixed integer then $ \Sigma a_n$ converges. Oh I see. So for part (a), by the comparison test I should prove that $ \Sigma a_n^p \geq \Sigma a_n$ and so we have convergence. $\endgroup$ – Nikitau Nov 9 '16 at 0:55
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1) Since $\sum_{n=1}^{\infty} a_n$ converges, we must have $\lim_n a_n=0$. Thus, there exists some $N$ such that $n \geq N \implies a_n < 1$. Moreover, $a_n^p < a_n$ for every $n \geq N$, since $p \geq 1$ and $0 \leq a_n < 1$. Therefore, $\sum_{n=1}^{\infty} a_n^p=\sum_{n=1}^{N-1} a_n^p + \sum_{n=N}^{\infty} a_n^p < \sum_{n=1}^{N-1} a_n^p+\sum_{n=N}^{\infty} a_n$ is finite, since $\sum_{n=1}^{N-1} a_n^p$ is a finite sum and $\sum_{n=1}^{\infty} a_n$ converges.

2) Just take $a_n = \frac{1}{n^{\frac{1}{p}}}$. Since $\frac{1}{p} > 1$, $\sum_{n=1}^{\infty} a_n$ converges by the integral test, but $\sum_{n=1}^{\infty} a_n^p=\sum_{n=1}^{\infty} \frac{1}{n}$ is the harmonic series, so it diverges.

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  • $\begingroup$ Thanks for the reply, but I'm a little confused over a few things. Wouldn't these terms $\sum_{n=1}^{N-1} a_n^p + \sum_{n=N}^{\infty} a_n^p < \sum_{n=1}^{N-1} a_n^p+\sum_{n=N}^{\infty} a_n$ be equal to each other? Or should the last part be written with small n? Also, for the last sentence did you mean to say that $\sum_{n=1}^{\infty} a_n^p$ converges? Why does showing that it's finite prove that it converges? $\endgroup$ – Nikitau Nov 9 '16 at 22:02
  • $\begingroup$ I am not sure I get the question, but that inequality comes from the fact that $a_n > a_n^p$ since $0 \leq a_n < 1$ and $p \geq 1$. As for the last sentence, what I am using is that something finite + something that converges gives some finite quantity. Convergence means finiteness of the limit of partial sums. $\endgroup$ – u1571372 Nov 9 '16 at 22:09
  • $\begingroup$ Thanks! It took me awhile but my book states it as convergence if the limit of the partial sums are bounded. Also, I had a typo for (iii). I want to show that $p < 1, \exists a_n$ such that $\Sigma a_n$ converges and $\Sigma (a_n)^p$ diverges. Would it work if I let $ a_n=\frac{1}{n\cdot(log n)^p}$? $\endgroup$ – Nikitau Nov 10 '16 at 20:45
  • $\begingroup$ The example I wrote in 2) is a sequence $a_n$ such that $\sum a_n$ converges but $\sum (a_n)^p$ diverges, just like you want it... $\endgroup$ – u1571372 Nov 10 '16 at 22:30

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