Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
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 1

up vote 4 down vote accepted

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 $a_n=2^{-n}$.

share|improve this answer
    
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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