Let $\sum_{n=1}^\infty a_n$ be a convergent series with $a_n \ge 0$. Prove if the following series converges. $$ \sum_{n=1}^\infty \frac{1}{n} \cdot \sqrt{a_n}$$

We have $\sum_{n=1}^\infty (\frac{1}{n} \cdot \sqrt{a_n}) = \sum_{n=1}^\infty \frac{a_n}{n^2}$ . Since $(a_n)\longrightarrow0$, there exists a $k \gt 0$ with $|a_n| \lt k$ $\forall n \in \mathbb N$.

Therefore $\sum_{n=1}^\infty \frac{a_n}{n^2} < \sum_{n=1}^\infty \frac{k}{n^2} = k\cdot\sum_{n=1}^\infty \frac{1}{n^2}$. So this series converges for all $a_n$.

Can I do it like this?


Does the following series converge? $$\sum_{n=1}^\infty (\sqrt[n]{a} -1)$$

Literally no ideas left how to get on this. I tried different ways but I didn't get very far; I hope you can help me out here.

  • $\begingroup$ Wait, what happened to the square roots?! That's very important to the problem! $\endgroup$ – Simply Beautiful Art Dec 16 '16 at 23:43
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    $\begingroup$ Why do we have $\sum_{n=1}^\infty\frac{\sqrt{a_n}}{n}=\sum_{n=1}^\infty\frac{a_n}{n^2}$?. $\endgroup$ – CIJ Dec 16 '16 at 23:44
  • $\begingroup$ There is a big leap on line 3. $\endgroup$ – copper.hat Dec 16 '16 at 23:46
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    $\begingroup$ In fact we only have $$\left(\sum_{n=1}^\infty \frac{\sqrt{a_n}}{n}\right)^2\leq \left(\sum_{n=1}^\infty \frac{1}{n^2}\right)\left(\sum_{n=1}^\infty a_n\right)$$ by Cauchy-Schwarz inequality. $\endgroup$ – Frank Lu Dec 16 '16 at 23:47
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    $\begingroup$ See math.stackexchange.com/questions/163441 $\endgroup$ – user84413 Dec 17 '16 at 20:58
  1. "We have $\sum_{k=1}^\infty \frac1n\sqrt{a_n}=\sum_{k=1}^\infty \frac{a_n}{n^2}$": do we? I'd say that, actually, the RHS is strictly smaller than the LHS for all sequences with $0<a_n<1$. Therefore, a lot of cases where your claim is false.

    The best way that comes to (not only my) mind is using Cauchy-Schwarz inequality $$\sum_{k=1}^\infty \frac1n\sqrt{a_n}\le\sqrt{\sum_{k=1}^\infty\frac1{n^2}}\cdot \sqrt{\sum_{k=1}^\infty a_n}<+\infty$$

  2. Hint: Notice that, for $a\ne 1$, $$\lim_{n\to\infty}\frac{\sqrt[n]{a}-1}{1/n}=\lim_{n\to\infty} n\left(-1+\exp\frac{\ln a}n\right)=\ln a\ne0$$ Therefore, the idea is that your series should behave roughly like a harmonic series.

  • $\begingroup$ Indeed it seems like i am not allowed to ^2. Unfortunately we didnt have the Cauchy-Schwarz inequality and the logarithmn yet. $\endgroup$ – user391105 Dec 16 '16 at 23:55
  • $\begingroup$ That's how I'd do it. ¯\ _(ツ)_/¯ Personally, I've learnt logarithms one year before being introduced to calculus and two years before working actively with series, so I wouldn't really know how to work that sort of machinery. $\endgroup$ – user228113 Dec 16 '16 at 23:57

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