Ch.3 #7. Prove that the convergence of $\sum a_n$ implies the convergence of $$\sum \frac{\sqrt{a_n}}{n},$$ if $a_n \geq 0$.

My attempt. If $\sum a_n$ is convergent, then by the root test, $\lim_{n \rightarrow \infty} sup \sqrt[n]{|a_n|} < 1$. This means $|a_n|^{1/n} < 1$. Because $a_n \geq 0$, this means $a_n ^{1/n} < 1$. Now apply the root test to $\sum \frac{\sqrt{a_n}}{n}$ by considering $$\lim_{n \to \infty} sup \sqrt[n]{|\frac{\sqrt{a_n}}{n}|}.$$ $$\sqrt[n]{|\frac{\sqrt{a_n}}{n}|} = \frac{\sqrt{{a_n}^{\frac{1}{n}}}}{\sqrt[n]n}.$$ Because $a_n ^{1/n} < 1$, certainly the numerator on the left-hand side of the equation above is $<1$. And because $n$ tends to $\infty$, and $n > 1 \rightarrow \sqrt[n]n > 1,$ the denominator is $>1$. Therefore the limit is $<1$ and the root test implies that $\sum \frac{\sqrt{a_n}}{n}$ converges.

If there are any errors in my proof, I think they come from 1) my assumption of the converse of the root test theorem and 2) my misunderstanding of $\lim_{n\to\infty} sup$. Thanks in advance for your help.

  • $\begingroup$ Well, you've identified the error all by yourself! You say something about your assumption of the converse of the root test. You know that when you're proving things you can't just assume stuff. The converse of the root test is not true (at least your version of the converse). Think about the Cauchy-Schwarz inequality... $\endgroup$ – David C. Ullrich Jul 7 '15 at 17:33
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    $\begingroup$ $\sum 1/n^2$ converges, but $\lim_n \root n\of {n^2}=1$. Instead, you could use the inequality $2ab\le (a^2+b^2)$ and the Comparison Test. $\endgroup$ – David Mitra Jul 7 '15 at 17:34
  • $\begingroup$ You're right, you can't assume the original limit is $< 1$, it could be $1$ (and yet you still have convergence) in which case your limit for the second series is also $1$ so you can't say anything about convergence using that test. $\endgroup$ – user2566092 Jul 7 '15 at 17:34
  • $\begingroup$ Your assumption of the converse i'snt right, the converse is \$\limsup_n \sqrt[n]{a_n} \leq 1$, you can use Cauchy-Schwarz to prove the convergence. $\endgroup$ – Euler88 ... Jul 7 '15 at 17:35
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    $\begingroup$ You are correct. The converse of the root theorem requires the non-strict inequality. That is, if $\sum a_n$ converges, then $\lim \sup a_n^{1/n} \le 1$. For example, $\sum n^{-2}$ converges, but $n^{-2/n}\to 1$. $\endgroup$ – Mark Viola Jul 7 '15 at 19:45

David Mitra basically spelled it out but in case you're unclear on the details, note $(a-b)^2 \geq 0$ gives you $ab \leq a^2/2 + b^2/2$. Then, set $a = \sqrt{a_n}$ and $b = 1/n$ and you will reduce proving convergence of your second series to simply proving that $\sum_n 1/n^2$ converges, for which you can use the Cauchy condensation test if you can't just assume convergence automatically for that series.

  • $\begingroup$ Thanks, that's a clever solution. Just out of curiosity, how do you and other people who answered find quick solutions like these? Have you seen the problem before, or does it just come with experience? $\endgroup$ – user217664 Jul 7 '15 at 17:47
  • $\begingroup$ @agrasin It definitely helps to be familiar with things like Cauchy-Schwarz inequality. But one intuitive way to see this proof idea is to realize that a sum of two series is often much easier to analyze for convergence compared to a series where terms are products. So if you know the inequality, it is a natural thing to think of since it reduces convergence of the series of products to the convergence of two separate series, one of which you already know convergence as an assumption. The fact that the square in the inequality directly cancels the given square-root is just happy coincidence. $\endgroup$ – user2566092 Jul 7 '15 at 17:51

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