# convergence and divergence of a series

Let {$a_n$} be a sequence of non-negative real numbers such that the series $\sum^\infty_{n=1} {a_n}$ is convergent. If $p$ is a real number such that the series $\sum^\infty_{n=1}\frac{\sqrt{a_n}}{n^p}$ diverges, then what can we say about the values of $p$ and how?

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You can say at least the trivial thing: $p\leq 1$ (since $a_n=o(1)$). – Clement C. May 8 '13 at 13:01
(also, if $(a_n)$ is nonincreasing, you will have $a_n=o(\frac 1 n)$, so to have the new series to diverge, you'd have to have $p\leq 1/2$) – Clement C. May 8 '13 at 13:11
Thank you, sir. But how can we conclude that? please explain. – Nimit May 8 '13 at 13:11
For which one? For the first: As $a_n = o(1)$, $\sqrt{a_n} = o(1)$. If $p > 1$, $0\leq \frac{\sqrt{a_n}}{n^p} = o\!\left(\frac{1}{n^p}\right)$, and thus the series $\sum\frac{\sqrt{a_n}}{n^p}$ would converge, as $\sum\frac{1}{n^p}$ does ($p>1$) — contradicting the hypothesis. – Clement C. May 8 '13 at 13:14

## 1 Answer

What we can say is "if such a $p$ exists, then $p\leqslant 1/2$". Otherwise, by Cauchy-Schwarz inequality, we would deduce the convergence of $\sum_n\frac{\sqrt{a_n}}{n^p}$.

But such a $p$ doesn't need to exist, for example when $a_n=2^{-n}$.

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One more comment, it is worth mentioning that $p\leq 1/2$ is achievable, for example for $a_n =\frac{1}{n \ln^2(n)}$. So the bound cannot be improved in general. – N. S. May 8 '13 at 13:32
As mentioned by @Clement C., $p$ can be less or equal to 1. I am a bit confused between $1/2$ and $1$. Could you please explain? – Nimit May 8 '13 at 13:32
@N.S.: if $a_n=\frac1{n\log(n)^2}$ then $p=1/2$ fails to converge. Ah, I see your comment changed :-) – robjohn May 8 '13 at 13:34
@gaathiyo This answer shows that the bound $p\leq 1$ is not optimal and can be improved.... Basically it is the following problem: if you have to find a property of a number $x$ and you decide that $x <1$, then saying that $x<1,000,000$ is also true, but not really the right answer. – N. S. May 8 '13 at 13:34