How to prove $\sum_{n=1}^{+\infty}\dfrac{1}{(n+1)\sqrt[e]{n}}$ converges to a number $I can prove that  $\sum_{n=1}^{+\infty}\dfrac{1}{(n+1)\sqrt[e]{n}}$ converges, but don't know how to transform the $\sqrt[e]{n}$ item to prove it less than $e$, could someone help/hint me?

We use $\int x^{-1-1/e} dx = -ex^{-1/e} + C$.
$$\sum_{n=1}^{\infty} \frac{(1+1/n)^{1/e}}{(n+1)^{1+1/e}} = \frac{1}{2} + \frac{1}{3 \times 2^{1/e}} +  \frac{1}{4 \times 3^{1/e}} + \sum_{n=4}^{\infty} \frac{(1+1/n)^{1/e}}{(n+1)^{1+1/e}} $$
$$  \le \frac{1}{2} + \frac{1}{3 \times 2^{1/e}} +  \frac{1}{4 \times 3^{1/e}} + (\frac{5}{4})^{1/e} \int_4^{\infty} x^{-1-1/e} dx $$
$$  = \frac{1}{2} + \frac{1}{3 \times 2^{1/e}} +  \frac{1}{4 \times 3^{1/e}} + (\frac{5}{16})^{1/e} e \approx 2.697177 < e$$
