Convergence of the series $\sum a^{\sum \frac{1}{n}}$ [duplicate]

I need to find the values of $a \in \mathbb{R}^+$ for which the following sum converges. $$\sum_{k=1}^\infty a^{\sum_{n=1}^k \frac{1}{n}}$$

We get, by ratio Test, $$\left|\frac{a^{\sum_{n=1}^{k+1} \frac{1}{n}}}{a^{\sum_{n=1}^k \frac{1}{n}}}\right| = a^{\frac{1}{k+1}} \geq 1 \text{ if } a\geq 1$$ and so the series diverges if $a\geq 1$.

But how do I handle the case for $0 <a < 1$ ?

P.S. $\mathop {\lim \sup }\limits_{n \to \infty } \left|\frac{a^{\sum_{n=1}^{k+1} \frac{1}{n}}}{a^{\sum_{n=1}^k \frac{1}{n}}}\right| = 1.$ So, that's inconclusive (by Theorem 3.34 of Baby Rudin.)

We compute \begin{align*} \mathop {\lim }\limits_{k \to + \infty } k\left( {\frac{{a^{\sum\nolimits_{n = 1}^k {\frac{1}{n}} } }}{{a^{\sum\nolimits_{n = 1}^{k + 1} {\frac{1}{n}} } }} - 1} \right) = \mathop {\lim }\limits_{k \to + \infty } k\left( {a^{ - \frac{1}{{k + 1}}} - 1} \right) = \mathop {\lim }\limits_{k \to + \infty } k\left( {e^{ - \frac{1}{{k + 1}}\log a} - 1} \right) = - \log a. \end{align*} By Raabe's test the series converges when $-\log a>1$, i.e., $0<a<e^{-1}$ and diverges when $a>e^{-1}$. To check the convergence for $a=e^{-1}$, we use $ \sum\nolimits_{n = 1}^k {\frac{1}{n}} <\log k +1$, to find $$ \sum\limits_{k = 1}^\infty {e^{ - \sum\nolimits_{n = 1}^k {\frac{1}{n}} } } \ge e^{ - 1} \sum\limits_{k = 1}^\infty {e^{ - \log k} } = e^{ - 1} \sum\limits_{k = 1}^\infty {\frac{1}{k}} . $$ But the harmonic series diverges, hence the series diverges when $a=e^{-1}$.