Prove that $\sum_{n=1}^{\infty}\left( 1-\frac{1}{n^2+1}\right)^{n^3}$ is convergent or divergent.

Question

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.

Answer 1

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.

Answer 2

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 $c>0$.

Answer 3

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.

Answer 4

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.