For given $\varepsilon>0$, I am trying to show that $\sum_{n=1}^\infty(1-\frac{1}{n+1}^{1-\varepsilon})^{n+1}$ is convergent. It is clear that comparison test is used.
Attempt: estimate using a geometric series. Let $R=\sup_n (1-1/(n+1)^{1-\varepsilon})$. Then since $(1-1/(n+1)^{1-\varepsilon})<1$ always we have that $R<1$ so $\sum R^{n+1}$ converges. The only issue is, apparently taking supremum always $R\le 1$ which is obviously bad as $R=1$ diverges.
We have that
$$\left(1-\frac{1}{(n+1)^{1-\varepsilon}}\right)^{n+1}=e^{(n+1)\log\left(1-\frac{1}{(n+1)^{1-\varepsilon}}\right)}\approx e^{(n+1)\left(-\frac{1}{(n+1)^{1-\varepsilon}}\right)}=e^{-(n+1)^{\varepsilon}}$$
then refer to limit comparison test.