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.
 A: Using $\log(1+x)\le x$ for $x\ge -1$, we have for $n\ge 1$
$$\begin{align}\left(1-\frac1{n^2+1}\right)^{n^3}&=e^{n^3 \log\left(1-\frac1{n^2+1}\right)}\\\\
&\le e^{-n^3/(n^2+1)}\\\\
&\le e^{-n/2}
\end{align}$$
And you can conclude now.
A: Hint: Use the inequality
$$
1-t\le e^{-t}.
$$
Now you can compare your series to the series
$$
\sum \exp\left(-\frac{n^3}{n^2+1}\right)
$$
which in turn can be compared to the series
$$
\sum e^{-cn}
$$
for an appropriate constant $c>0$.
A: Render
$(1-\frac{1}{n^2+1})^{n^3}=(\frac{n^2}{n^2+1})^{n^3}=(1+\frac{1}{n^2})^{-n^3}$ (**)
Then as $n\to\infty, (1+\frac{1}{n^2})^{-n^\color{blue}{2}}\to1/e$, so the terms of the series $\in O(2^{-n})$.  Therefore convergence by comparison with a geometric series.
(**)  You may have missed the minus sign in the exponent when you set up your comparison test.
A: 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.
