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!

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.