Let $\alpha,\beta,c>0$. Then I have the strong assumption that
$$\sum_{n=1}^\infty n^\alpha \exp(-cn^\beta)$$
converges. How would you show that? Using the root test you get only convergence for $\beta>1$ (otherwise the limit is $1$) if I'm not mistaken.
Using the ratio test it might work. For that you would have to show that
$$
\lim_{n \to \infty} \frac{(n+1)^\alpha}{n^\alpha} = 1
$$
and
$$
\limsup_{n \to \infty} (n+1)^\beta-n^\beta>0.
$$
Best Answer
For all integers $k \ge 0$ and real numbers $x \ge 0$ is $$ e^x \ge \frac{x^k}{k!} \, . $$ It follows that $$ n^\alpha \exp(-cn^\beta) = \frac{n^\alpha}{\exp(cn^\beta)} \le \frac{k! }{c^k } n^{\alpha-k\beta} \, . $$ Since $\beta$ is positive we can choose $k$ so large that $\alpha-k\beta < -1$, this shows that the series is convergent.