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Yesterday I stumbled across an interesting exercise (Indam test 2014, Exercise B3):

(Ex) Given a positive sequence $\{a_n\}_{n\geq 1}$ such that $\sum_{n\geq 1}a_n$ is convergent, prove that $$ \sum_{n\geq 1}a_n^{\frac{n-1}{n}}$$ is convergent, too.

My proof exploits an idea from Carleman's inequality. We have: $$ a_n^{\frac{n-1}{n}}=\text{GM}\left(\frac{1}{n},2a_n,\frac{3}{2}a_n,\ldots,\frac{n}{n-1}a_n\right) $$ and by the AM-GM inequality $$ a_n^{\frac{n-1}{n}}\leq \frac{1}{n}\left(\frac{1}{n}+a_n\sum_{k=1}^{n-1}\frac{k+1}{k}\right)\leq \frac{1}{n^2}+\left(1+\frac{\log n}{n}\right)a_n $$ hence $$ \sum_{n\geq 1}a_n^{\frac{n-1}{n}}\color{red}{\leq} \frac{\pi^2}{6}+\left(1+\frac{1}{e}\right)\sum_{n\geq 1}a_n.$$

Now my actual

Question: Is there a simpler proof of (Ex), maybe through Holder's inequality, maybe exploiting the approximations $$ \sum_{m<n\leq 2m}a_n^{\frac{2m-1}{2m}}\approx \sum_{m<n\leq 2m}a_n^{\frac{n-1}{n}}\approx \sum_{m<n\leq 2m}a_n^{\frac{m-1}{m}}$$ "blocking" the exponents over small summation sub-ranges?

Mittens
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Jack D'Aurizio
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    It might be useful to exploit the trivial fact that $A=\left{n\in\mathbb{N}^*: a_n\geq 1\right}$ is a finite set, in order to "reverse" the trivial inequality $$\forall x,\alpha\in(0,1),\qquad x^{\alpha}\color{red}{\geq} x. $$ – Jack D'Aurizio Oct 01 '16 at 12:02
  • Isn't it enough to exploit Euler-MacLaurin summation formula

    $$ S=\sum_{n\geq1}a_n^{1-1/n}\sim \int_{1}^{\infty}a(x)^{1-1/x}=\int_{1}^{\infty}a(x)e^{\frac{1}{x}\log(a(x))}=\int_{1}^{\infty}a(x)\sum_{m\geq0}\frac{1}{m!}\frac{a(x)^m}{x^m}\sim\int_{1}^{\infty}a(x)=C $$

    by the fact that $a(x)/x<a(x)$ on $x\in(1,\infty)$?

     – tired Oct 15 '16 at 01:01

4 Answers4

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Define $$S=\{n | a_n \leq \frac{1}{2^n}\}$$ $$T=\{n | \frac{1}{2^n} < a_n\}$$

Since $a_n$ is positive it suffices to show that $$\sum\limits_{n\in S} \frac{a_n}{\sqrt[n]{a_n}} \ \text{and} \ \ \sum\limits_{n\in T} \frac{a_n}{\sqrt[n]{a_n}} $$ converge separately.

If $n\in S$ then $$a_n^{\frac{n-1}{n}} \leq \frac{1}{2^{n-1}}$$ thus the first series converges.

If $n \in T$ then $$\frac{1}{2}<\sqrt[n]{a_n}$$ thus $$\frac{a_n}{\sqrt[n]{a_n}}<2a_n$$

ad the second series will also converge by comparison with $\sum\limits_{n\in T}2a_n$

Rene Schipperus
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    (+1) Oh, I love this approach. So simple, so neat. But I am waiting for further opinions to come before accepting your answer, I hope you do not mind. – Jack D'Aurizio Oct 01 '16 at 12:48
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    Thanks. I dont mind, quite the contrary, the details need to be checked. – Rene Schipperus Oct 01 '16 at 12:50
  • @Michael I agree the underlying principle is the same. – Rene Schipperus Oct 01 '16 at 13:11
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    It is interesting to point out that by replacing the constant $2$ above with a generic $\alpha > 1$, we get the inequality $$\sum_{n\geq 1}a_n^{\frac{n-1}{n}}\leq \frac{\alpha}{\alpha-1}+\alpha\sum_{n\geq 1}a_n $$ and by minimizing the RHS on $\alpha$ we get $$ \sum_{n\geq 1}a_n^{\frac{n-1}{n}}\leq \left(1+\sqrt{\sum_{n\geq 1}a_n}\right)^2 $$ !!! – Jack D'Aurizio Oct 01 '16 at 13:38
  • @Michael Personally, I dont see it as necessary to delete your answer, the different exposition may give some people more insight into what is going on. But it is up to you of course. – Rene Schipperus Oct 02 '16 at 00:49
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    @Michael Yeah, I have the same problem all the time, I dont even try to answer elementary calculus questions anymore, because I know as soon as I type an answer there will be three others basically the same as mine. – Rene Schipperus Oct 02 '16 at 01:01
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My friend came up with the following solution. The proof is very similar to OP's one. It is a bit easier as it does not involve facts about harmonic numbers, logarithms and $e$; however the bound provided by OP is tighter.

AM-GM yields $$a_n^{(n-1)/n} = \sqrt[n]{\frac 1n \cdot na_n \cdot \underbrace{a_n \cdot \ldots \cdot a_n}_{n-2 \text{ times}}} < \frac 1n \cdot \left(\frac 1n + na_n + \underbrace{a_n + \ldots + a_n}_{n-2 \text{ times}}\right) < \frac {1}{n^2} + 2a_n.$$

Thus $$\sum_{n=1}^\infty a_n^{(n-1)/n} < \sum_{n=1}^\infty \frac 1{n^2} + 2\sum_{n=1}^\infty a_n < \infty.$$

timon92
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Since $\sum_{n\ge 1}a_n$ is convergent, you can say that $a_n \approx_{n\to \infty} u_n$ with $u_n < {1\over n}$. Raising this inequality to to the power ${n-1}\over n$ you get :

$$ a_n^{{n-1}\over n}\approx_{n\to \infty} u_n^{{n-1}\over n} < {1 \over n^{1+{{n-1}\over n}}} = b_n $$

EDIT : The previous inequlity is wrong, here is the correct one : $$ a_n^{{n-1}\over n}\approx_{n\to \infty} u_n^{{n-1}\over n} < {1 \over n^{{{n-1}\over n}}} = b_n $$ Unfortunatly this does not provide enough with regard to Riemann rule to conclude to anything.

With $b_n$ convergent too by Riemann. Since all those sequences are positives, you can deduce $\sum_{n\ge 1}a_n^{{n-1}\over n}$ converge too.

Furrane
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  • (+1) This is an interesting approach, too. It would be interesting to provide an explicit bound depending on $K=\sup_{n\to +\infty}na_n$, at this point. – Jack D'Aurizio Oct 01 '16 at 13:05
  • However, by raising $u_n< \frac{1}{n}$ to the power $\frac{n-1}{n}$ I only get $u_n^{\frac{n-1}{n}}<\frac{1}{n^{\frac{n-1}{n}}}$, and $\sum_{n\geq 1}n^{\frac{1-n}{n}}$ is not converging. – Jack D'Aurizio Oct 01 '16 at 13:09
  • The problem is that you can have $\sum_{n=1}^{\infty}a_n$ convergent, yet still have $a_n = 1/\sqrt{n}$ for infinitely many $n$. So the approximation $a_n \approx u_n$ with $u_n \leq 1/n$ does not necessarily hold for all sufficiently large $n$. – Michael Oct 01 '16 at 13:24
  • Riemann tells us that $\sum_1^\infty {1\over n^\alpha}$ converge for $\alpha > 1$, and if $\alpha \le 1$ it "blows" to $+\infty$ . In your example Michael $\sum_1^\infty {1\over n^{1/2}} = \sum_1^\infty {1\over n^\alpha}$ with $\alpha=1/2$ so It does NOT converge. – Furrane Oct 01 '16 at 14:28
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    Jack you're absolutly right, I did a terrible mistake : $(x^a)^b = x^{ab} \ne x^{a+b}$. – Furrane Oct 01 '16 at 14:33
  • @Furrane : No. Consider $a_n = 1/\sqrt{n}$ for those indices $n$ which are powers of 2, and $a_n=0$ if $n$ is not a power of 2. Then indeed $a_n=1/\sqrt{n}$ for infinitely many $n$, but: $$\sum_{n=1}^{\infty} a_n = \sum_{k=1}^{\infty} \frac{1}{\sqrt{2^k}} = 1+\sqrt{2} < \infty $$ – Michael Oct 01 '16 at 21:09
  • Michael you have to understand that in the Riemann rule $\alpha$ is a constant. I refere you the demonstration of Henry on this post http://math.stackexchange.com/questions/235574/convergence-of-sum-n-1-infty-frac1n-alpha – Furrane Oct 01 '16 at 21:18
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    @Furrane : Not all sequences have the form $a_n = 1/n^{\alpha}$ for all $n$. If a sequence has that form, clearly finite-summability is determined by $\alpha$. Yet, this question asked about general sequences. My example shows that if you take the sequence ${1/\sqrt{n}}_{n=1}^{\infty}$, which obviously diverges to $\infty$ when summed, and change some of the terms to $0$ (while leaving an infinite number of nonzero terms), the result can converge to $1+\sqrt{2}$. So, the statement "$a_n \leq 1/n$ for suffficiently large $n$" is false. So, the first sentence of your answer is incorrect. – Michael Oct 02 '16 at 00:27
  • @JackD'Aurizio : In particular, in regards to your first comment, it is possible for $K= \lim_{k\rightarrow\infty}\sup_{n\geq k}|na_n|=\infty$ even when $\sum_{n=1}^{\infty}|a_n|<\infty$. – Michael Oct 02 '16 at 00:36
  • @Michael: you are clearly right. I realized that few seconds after writing my initial comment, but it is faulty quite like the outlined approach :/ – Jack D'Aurizio Oct 02 '16 at 00:42
  • I see my mistake now Michael. So my answer was based on a wrong fact and eventually would lead to a useless result. Wow I guess the silver lining i I can only go up from there :) Thanks for your patience to explain to me even if I though I was right for too long. – Furrane Oct 02 '16 at 07:26
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This is to expand on a comment by Jack D'Aurizo and to see that a similar inequality holds for integrals over $[0,\infty)$. To be more precise:

Suppose $f$ is a positive function on $[1,\infty)$ ad that $f\in L_1([1,\infty)$. Then $$\int^\infty_1\big(f(x)\big)^{1-\frac1x}\,dx<\infty.$$

Let $\lambda>1$. Splitting the las integral we have \begin{align} \int^\infty_1\big(f(x)\big)^{1-\frac1x}\,dx=\int_{\{x\geq1:f(x)\leq \lambda^{-x}\}} \big(f(x)\big)^{1-\frac1x}\,dx+\int_{\{x\geq1: f(x)>\lambda^{-x}\}}\big(f(x)\big)^{1-\frac1x}\,dx\tag{0}\label{zero} \end{align}

The first integral on the right-hand-side of \eqref{zero} is bounded as follows: $$ \int_{\{x\geq1:f(x)\leq\lambda^{-x}\}} \big(f(x)\big)^{1-\frac1x}\,dx<\int^\infty_1\lambda^{1-x}\,dx=\frac{1}{\log\lambda} $$ The second integral on the right-hand-side of \eqref{zero} is bounded by $$ \int_{\{x\geq1:f(x)> \lambda^{-x}\}} \big(f(x)\big)^{1-\frac1x}\,dx<\lambda\int^\infty_1 f(x)\,dx $$ hence

\begin{align} \int^\infty_1\big(f(x)\big)^{1-\frac1x}\,dx\leq \lambda\Big(\int^\infty_1f\Big) +\frac{1}{\log\lambda}\leq \lambda\Big(\int^\infty_1f\Big)+\frac{\lambda}{\lambda-1}\tag{1}\label{one} \end{align} where the last inequality follows from $\frac{t}{1+t}\leq \log(1+t)\leq t$ for all $t>-1$. Minimizing the right-hand-side of \eqref{one} yields $$\int^\infty_1\big(f(x)\big)^{1-\frac1x}\,dx\leq \left(1+\sqrt{\int^\infty_1f(x)\,dx}\right)^2$$

Riemann
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Mittens
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