# Is $\sum \frac{1}{a^{1+{1\over a}}}$ convergent? [duplicate]

Is $\sum \frac{1}{a^{1+{1\over a}}}$ convergent? I know some fact that if $\sum \frac{1}{a^s}$ is convergent for any $s>1$ but here, the power is varying, more precies it is decreasing to 1 but i don't know whether it is convergent or divergent and how to prove to test or to to prove that it is divergent.

## marked as duplicate by Thomas, Alexander Gruber♦, Davide Giraudo, draks ..., rschwiebMar 7 '13 at 15:22

• Do you mean $\sum_{a=1}^\infty \dfrac{1}{a^{1+1/a}}$? – Siméon Mar 7 '13 at 11:48
• @Ju'x YES, THAT'S WHAT I MEAN – AAA Mar 7 '13 at 11:53
• no need to yell...the only excuse (I can think of) might be a broken caps lock! – draks ... Mar 7 '13 at 15:17

It is easy to show that if $a$ is a positive integer, then $a^{1/a}\lt 2$. For this is equivalent to showing that $a\lt 2^a$, which can be done by induction or in other ways.
Thus the $a$-th term of our series is $\gt \frac{1}{2a}$, so our series diverges by Comparison with the harmonic series.
Using the inequality $\ln(u) \leq u$, we have the lower bound $$\frac{1}{n^{1+1/n}} = \frac{1}{n}\exp(-\frac{\ln n}{n}) \geq \frac{1}{e}\frac{1}{n}$$
Hence by comparison, $$\sum_{n=1}^\infty \frac{1}{n^{1+1/n}} = +\infty.$$
Compare with the harmonic series $\displaystyle \sum_{a=1}^\infty\dfrac1a$ $\left(\displaystyle \lim_{a\to\infty}\dfrac{1}{a^{1+\frac1a}}/\dfrac1a=\ldots\right)$.