<s>[INST] For what positive numbers $a,b$ is the series $\sum_{k=0}^{\infty} a^{\frac{1}{b}+\frac{1}{b+1}+...+\frac{1}{b+k}}$ convergent? Last one for tonight. For what positive numbers $a,b$ is the series $$\sum_{k=0}^{\infty} a^{\frac{1}{b}+\frac{1}{b+1}+...+\frac{1}{b+k}}$$ convergent?
I'm defeated by this one. I don't even know how to begin.
 [/INST] Because$$\def\d{\mathrm{d}}
\sum_{j = 0}^k \frac{1}{b + j} > \int_0^{k + 1} \frac{\d x}{b + x} = \ln(b + k + 1) - \ln b,
$$
for $a \geqslant 1$,$$
a^{\sum\limits_{j = 0}^k \frac{1}{b + j}} \geqslant a^{\ln(b + k + 1) - \ln b} \geqslant a^{\ln(b + 1) - \ln b},
$$
which implies $a^{\sum\limits_{j = 0}^k \frac{1}{b + j}} \not\to 0 \ (k \to \infty)$. Therefore the series is divergent.
For $0 < a < \mathrm{e}^{-1}$,\begin{align*}
\sum_{k = 0}^\infty a^{\sum\limits_{j = 0}^k \frac{1}{b + j}} &\leqslant \sum_{k = 0}^\infty a^{\ln(b + k + 1) - \ln b} = a^{-\ln b} \sum_{k = 0}^\infty a^{\ln(b + k + 1)}\\
&= a^{-\ln b} \sum_{k = 0}^\infty (\mathrm{e}^{\ln a})^{\ln(b + k + 1)} = a^{-\ln b} \sum_{k = 0}^\infty (\mathrm{e}^{\ln(b + k + 1)})^{\ln a} \\
&= a^{-\ln b} \sum_{k = 0}^\infty (b + k + 1)^{\ln a} < +\infty.
\end{align*}
For $\mathrm{e}^{-1} \leqslant a < 1$, because$$
\sum_{j = 0}^k \frac{1}{b + j} = \frac{1}{b} + \sum_{j = 1}^k \frac{1}{b + j} \leqslant \frac{1}{b} + \int_0^k \frac{\d x}{b + x} = \frac{1}{b} + \ln(b + k) - \ln b,
$$
then analogously,\begin{align*}
\sum_{k = 0}^\infty a^{\sum\limits_{j = 0}^k \frac{1}{b + j}} &\geqslant \sum_{k = 0}^\infty a^{\frac{1}{b} + \ln(b + k) - \ln b} = a^{\frac{1}{b} - \ln b} \sum_{k = 0}^\infty (b + k)^{\ln a} \geqslant a^{\frac{1}{b} - \ln b} \sum_{k = 0}^\infty \frac{1}{b + k} = +\infty.
\end{align*}
</s>