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I really don't want spoilers for this problem, but I am wondering if my approach is correct. (I am of course looking for a solution based on "standard" stuff regarding series, not relying on some exotic theorem from which it may follow immediately.)

Let $ a_n > 0 $ for all $ n = 1, 2, 3, \ldots$ and suppose $ \sum\limits_{n=1}^{\infty} a_n $ converges. Let $ \sigma_n = \frac{n}{n+1} $.

To show that $ \sum\limits_{n=1}^{\infty} a_n^{\sigma_n} $ converges, first observe that

$$ \limsup\limits_{n \to \infty} ([a_n^{\sigma_n}]^{1/n}) = \limsup\limits_{n \to \infty} (a_n^{1/(n+1)}) = \limsup\limits_{n \to \infty} (a_n^{1/n}) $$

so since $ \sum a_n $ converges, $ \limsup \sqrt[n]{a_n^{\sigma_n}} \leq 1 $.

If $ \limsup (\sqrt[n]{a_n^{\sigma_n}}) < 1 $, the result follows (by the Root test). Suppose $ \limsup (\sqrt[n]{a_n^{\sigma_n}}) = 1 $ intsead. Then (this part may be totally wrong) since $ a_n > 0 $ and $ a_n \to 0 $, $ \lim\limits_{n \to \infty} \sqrt[n]{a_n^{\sigma_n}} = 1 $, which means

$$ \lim\limits_{n \to \infty} \dfrac{1}{\sqrt[n]{a_n}} = \liminf\limits_{n \to \infty} \dfrac{1}{\sqrt[n]{a_n}} = 1. $$

Hence one can find $ N \in \mathbf{N} $ such that for all $ n \geq N $, (i) $ 0 < a_n < 1 $ and (ii) $ 1 < \dfrac{1}{a_n^{1/(n+1)}} < 2 $. Then

$$ |a_n^{\sigma_n}| = \left|a_n^{1 - \tfrac{1}{n+1}}\right| = |a_n| \cdot \left|\dfrac{1}{a_n^{1/(n+1)}}\right| \leq 2|a_n| $$

so $ \sum\limits_{n=1}^{\infty} a_n^{\sigma_n} $ must converge by the Comparison test.

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  • $\begingroup$ First, $\limsup$ implies nothing about $\lim$ (indeed, you might just sprinkle in some (almost) zero terms in the original series $\endgroup$ Apr 7, 2018 at 10:17
  • $\begingroup$ That makes sense. I guess was under the impression that $ a_n > 0 $ guaranteed they were the same, which is wrong. I'm wondering if the Root test actually helps at all with this problem... $\endgroup$
    – dasaphro
    Apr 7, 2018 at 10:22
  • $\begingroup$ This (or a very similar variation) has been asked many times before here... $\endgroup$ Apr 7, 2018 at 10:45
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    $\begingroup$ I don't think the root test helps at all. It looks like you are trying to extend the root test to the case when the test gives a result of 1...if it were possible to do that, it would be a standard part of the root test. You need the (possibly exotic) Holder inequality. $\endgroup$
    – Michael
    Apr 7, 2018 at 11:23
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    $\begingroup$ The idea is considering case by case. When the sequence is too small, and otherwise. $\endgroup$ Apr 7, 2018 at 18:38

2 Answers 2

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Surely, $a_n$ must be all positive. Then, we have the following inequality for all $n > 0$:

$a_n^{\sigma_n} \le \text{max} \left( \frac{1}{2^{n+1}}, a_n \right)^{\sigma_n} \le 2 \cdot \text{max}\left( \frac{1}{2^{n+1}}, a_n \right)$

Which leads directly to our the desired conclusion.

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  • $\begingroup$ Excellent!.................. $\endgroup$ Apr 8, 2018 at 8:19
  • $\begingroup$ So good. +1 ... $\endgroup$
    – Hans
    May 22, 2018 at 5:14
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Re the "this part may be totally wrong", yes it is, if you consider $a_n=\frac{1}{e^{n}}$ where $\sum\limits_{n}a_n=\frac{e}{e-1}$. Since $$\sqrt[n]{a_n^{\sigma_n}}=\frac{1}{e^{\sigma_n}}\rightarrow\frac{1}{e}\ne 1$$


Incomplete work. I will use this statement where $b_n=a_n^{\sigma_n}$. Now, let's look at $$\frac{a_n}{b_n}=a_n^{1-\sigma_n}=a_n^{\frac{1}{n+1}} \tag{1}$$

If $\color{red}{\lim\limits_{n\rightarrow\infty}a_n^{\frac{1}{n+1}}=l >0}$, from the statement $\Rightarrow \sum\limits_{n}a_n^{\sigma_n}<\infty$


If $\color{red}{\lim\limits_{n\rightarrow\infty}a_n^{\frac{1}{n+1}}=0}$ then $$0<a_n^{\frac{1}{n+1}}<\varepsilon<1$$ from some $N$ onwards, then $$0<a_n^{\sigma_n}=a_n^{\frac{n}{n+1}}<\varepsilon^n<1$$ and $$0<\sum\limits_{n}a_n^{\sigma_n}= \sum\limits_{n=0}^{N}a_n^{\sigma_n}+\sum\limits_{n=N+1}a_n^{\sigma_n}<\sum\limits_{n=0}^{N}a_n^{\sigma_n}+\sum\limits_{n=N+1}\varepsilon^n=\\ \sum\limits_{n=0}^{N}a_n^{\sigma_n}+\varepsilon^{N+1}\sum\limits_{n=0}^{\infty}\varepsilon^n= \sum\limits_{n=0}^{N}a_n^{\sigma_n}+\frac{\varepsilon^{N+1}}{1-\varepsilon}<\\ \sum\limits_{n=0}^{N}a_n^{\sigma_n}+\frac{1}{1-\varepsilon}<\infty$$


And the last case $\color{red}{\lim\limits_{n\rightarrow\infty}a_n^{\frac{1}{n+1}}}$ doesn't exists ... (tbc)

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    $\begingroup$ The last possible option seems to be $\lim_{n\rightarrow\infty} a_n^{1/(n+1)}$ does not exist, or the limit equal zero. $\endgroup$ Apr 7, 2018 at 18:37
  • $\begingroup$ Yep, true. Removed the "last possible option". Thank you for pointing out! $\endgroup$
    – rtybase
    Apr 7, 2018 at 20:55
  • $\begingroup$ Yeah, I don't see how this would work considering the case when the limit does not exist. Also, if the limit is 0, it is equal to $ \limsup \sqrt[n]{a_n^{\sigma_n}} $ so your inequalities don't seem necessary... $\endgroup$
    – dasaphro
    Apr 8, 2018 at 10:05
  • $\begingroup$ My idea was to prove by contradiction that the limit exists, but Taro came with an easier answer in the meantime $\endgroup$
    – rtybase
    Apr 8, 2018 at 10:08

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