Is $\sum \frac{1}{a^{1+{1\over a}}}$ convergent?

Question

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.

Answer 1

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.

Answer 2

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, $$ \sum_{n=1}^\infty \frac{1}{n^{1+1/n}} = +\infty. $$

Answer 3

Compare with the harmonic series $\displaystyle \sum_{a=1}^\infty\dfrac1a$ $\left(\displaystyle \lim_{a\to\infty}\dfrac{1}{a^{1+\frac1a}}/\dfrac1a=\ldots\right)$.