# 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.

• How about proving $$\left( 1-\frac{1}{(n^2+1)}\right)^{n^3} \sim e^{-n}$$ Commented Mar 4, 2021 at 17:14
• Try using the Root Test. Commented Mar 4, 2021 at 17:23
• and then the series will be convergent? how can you prove this? Commented Mar 4, 2021 at 17:24
• tutorial.math.lamar.edu/classes/calcii/roottest.aspx Commented Mar 4, 2021 at 17:28
• I know what the root test is, I don't know how it helps me to solve this. If I use the root test here assuming that $\sum_{n=1}^{\infty}( 1-\frac{1}{(n^2+1)})^{n^3} \leq \sum_{n=1}^{\infty}( 1-\frac{1}{(n^2+1)})^{n}$ and then using the root test on the right side, I will get 1, which is inconclusive with the root test. Commented Mar 4, 2021 at 17:31

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.

• thank you very much. is there anyway to solve this without using the equality $e^{log(x)}=x$? Commented Mar 4, 2021 at 17:40
• @Somuser That equality is elementary from high school: it follows at once from the very definition of the logarithm and exponential functions ... Commented Mar 4, 2021 at 17:50
• I know this equality but I wouldn't think to use this "trick" I would keep trying using the convergence tests Commented Mar 4, 2021 at 17:56
• @Somuser Sure. Note that $\left(1+\frac1{n^2+1}\right)^{n^3}=\frac1{\left(1+\frac1{n^2}\right)^{n^3}}$. From Taylor's theorem $\left(1+\frac1{n^2}\right)^{n^3}=1+n+O(n^2)$. and now you can compare to the series with general terms $\frac1{n^2}$. Commented Mar 4, 2021 at 18:04

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 $$c>0$$.

• Looks similar to mine Commented Mar 4, 2021 at 17:34
• Yep, my $1-t\le e^{-t}$ is analogous to your $\log(1+x)\le x$. Commented Mar 4, 2021 at 17:39

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.

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.

• Your first line is wrong. It should be $\left( 1-\frac{1}{(n^2+1)}\right)^{n^3}= \frac{1}{\left( 1+\frac{1}{n^2}\right)^{n^3}}$. Commented Mar 4, 2021 at 17:45
• @TonyK Thank you fixed it. I replaced $n^2+1$ by $(n+1)^2$. Commented Mar 4, 2021 at 17:49