# An interesting exercise about converging positive series, involving $\sum_{n\geq 1}a_n^{\frac{n-1}{n}}$

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?

• 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 1 '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 1:01

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 $\sum\limits_{n\in T}2a_n$

• (+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 1 '16 at 12:48
• Thanks. I dont mind, quite the contrary, the details need to be checked. – Rene Schipperus Oct 1 '16 at 12:50
• @Michael I agree the underlying principle is the same. – Rene Schipperus Oct 1 '16 at 13:11
• 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 1 '16 at 13:38
• @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 2 '16 at 1:01

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

• Clever use of AM-GM, nice one – Gabriel Romon Aug 22 '18 at 18:56

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.

• (+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 1 '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 1 '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 1 '16 at 13:24
• Jack you're absolutly right, I did a terrible mistake : $(x^a)^b = x^{ab} \ne x^{a+b}$. – Furrane Oct 1 '16 at 14:33
• @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 2 '16 at 0:27