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?

Solution 1:

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

Solution 2:

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

Solution 3:

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.