$\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^{sinh(n)}}$
Because is $\displaystyle \sinh 1 = 1.1752011936438...$ and the function $\displaystyle \sinh (*)$ is [strongly] increasing, for all n will be...
$\displaystyle \frac{1}{n^{\sinh n}} < \frac{1}{n^{1.1}}$ (1)
Now the series...
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{1.1}}$ (2)
... converges so that...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$