Prove that $\sum_{n=1}^{\infty}\left( 1-\frac{1}{n^2+1}\right)^{n^3}$ is convergent or divergent. I have the series $$\sum_{n=1}^{\infty}\left( 1-\frac{1}{n^2+1}\right)^{n^3}.$$ How can I prove the convergence or divergence of it?
I tried to use the comparison test and claim that:
$$\sum_{n=1}^{\infty}\left( 1-\frac{1}{n^2+1}\right)^{n^3}\leq \sum_{n=1}^{\infty}\left( 1+\frac{1}{n^2}\right)^{n^2}$$ but the right series divergent because the limit of the sequence is e and therefore the series is divergent.

Hint
$$
\left( 1-\frac{1}{(n^2+1)}\right)^{n^3}= \frac{1}{\left( 1+\frac{1}{n^2}\right)^{n^3}}
$$
Now, for $n \geq 2$ you have
$$
\left( 1+\frac{1}{n^2}\right)^{n^3} \geq \binom{n^3}{2}(\frac{1}{n^2})^2=\frac{n^3-1}{2n}
$$
Therefore
$$
\sum_{n=1}^\infty \left( 1-\frac{1}{(n^2+1)}\right)^{n^3} \leq \frac{1}{2}+\sum_{n=2}^\infty \frac{2n}{n^3-1}
$$
It is trivial to show that the sum on the RHS is convergent.
