# What can be said about $\sum a_n^{\frac{n-1}{n}}$? [duplicate]

If $$a_n$$ are positive real numbers such that $$\sum a_n$$ converges, what can be said about $$\sum a_n^{\frac{n-1}{n}} ?$$

To see how the land lies let's test a few standard cases.

• If we have a converging geometric series $$a_n=aq^n, 0, $$a>0$$, then for all $$n\ge1$$ $$a_n^{(n-1)/n}=a^{(n-1)/n}q^{n-1}\le Aq^{n-1}$$ with $$A=\max\{1,a\}$$. Meaning that the new series is majorized by a converging geometric series.
• If $$a_n=1/n^{1+\epsilon}$$ for some $$\epsilon>0$$, another standard converging test series, then, for $$n>2/\epsilon$$, we have $$(n-1)/n>1-\dfrac\epsilon2$$, and consequently $$a_n^{(n-1)/n}<\frac1{n^{(1+\epsilon)(1-\epsilon/2)}}.$$ Meaning that we have again a converging majorant as the exponent is a constant $$>1$$.

But what happens in general?

• Unless I'm grossly mistaken, if $|a_n|>1$ for all $n$, then the second series is convergent by the comparison test, at the very least. Commented Aug 29, 2019 at 1:03
• Commented Oct 26, 2019 at 14:52

Hint: Let \begin{align} A &= \{n\in\mathbb{N}: a_n<2^{-n}\}\\ B &= \{n\in\mathbb{N}: a_n\ge 2^{-n}\} \end{align} (notice that $A\cup B = \mathbb{N}$). Then $$\sum\limits_{n\in A}{a_n^{\frac{n-1}{n}}}\le\sum\limits_{n\in A}{2^{-(n-1)}}$$ while for $n\in B$, we have $a_n\ge 2^{-n}\implies a_n^{-1/n}\le 2$, and hence $$\sum\limits_{n\in B}{a_n^{\frac{n-1}{n}}}\le\sum\limits_{n\in B}{2a_n}.$$

$$\sum a_n^{{(n-1)}/n}$$

$$\frac{a_{n+1}^{{n}/{(n+1)}}}{a_n^{{(n-1)}/n}}=\dfrac{\Big(\frac{a_{n+1}}{a_n}\Big)}{\Big(\frac{\sqrt[n+1]{a_{n+1}}}{\sqrt[n]{a_n}}\Big)}$$

Now let by root test $0<\lim_{n\to\infty}\sqrt[n]{a_n}=k< 1$, then also we have ratio test $0<\lim_{n\to\infty}\Big(\frac{a_{n+1}}{a_n}\Big)=l< 1$

Hence $$\lim_{n\to\infty}\frac{a_{n+1}^{{n}/{(n+1)}}}{a_n^{{(n-1)}/n}}=\dfrac{\Big(\frac{a_{n+1}}{a_n}\Big)}{\Big(\frac{\sqrt[n+1]{a_{n+1}}}{\sqrt[n]{a_n}}\Big)}=\dfrac{l}{\frac{k}{k}}=l<1$$

Hence the series converges.

If $\lim_{n\to\infty}\sqrt[n]{a_n}=1$, then also $\lim_{n\to\infty}\Big(\frac{a_{n+1}}{a_n}\Big)=1$, then we can not conclude in this way.

In this case we can find a large $N,m_n\in \mathbb{N}$, $a_n\sim \dfrac{1}{{m_n}^\alpha}$ for $n\geq N$, where $\alpha >1$. Then $a_n^{{(n-1)}/n}\sim \dfrac{1}{{m_n}^{\alpha(n-1)/n}}$. We can find $N_1\in\mathbb{N}$ s.t. $\alpha(n-1)/n>1$, for $n\geq N_1$. Hence the series converges.

Note: I am not sure if $k=l$(need to prove). But not needed here.