$\sum\left(\frac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} $ converges or diverges? The original question is to show that $\;\sum\left(\dfrac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} $ either converges or diverges.
I know it diverges but I'm having difficulty arriving at something useful for $  a_n $.
Here's what I did:
$  a_n = \left(\dfrac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} \geq\left( \dfrac{ n^2 + 1}{n^2 +2n + 1}\right)^{n^2}  \geq\left( \dfrac{ n^2 - 1}{n^2 +2n + 1}\right)^{n^2} = \left(\dfrac{(n-1)(n+1)}{(n+1)^2}\right)^{n^2}  = \left(\dfrac{n-1}{n+1}\right)^{n^2} = \left( 1 - \dfrac{2}{n+1}\right)^{n^2} $
Do you have idea how to continue from here? ( my idea was to arrive to some expression that goes to infinity and deduce by the comparison test that $ a_n $ diverges ). I feel like something is right infront of my eyes but I can't see it.
Edit: The series converges. However, how do you propose I should continue from where I've left?

Let's use the root test:
$$\sqrt[\Large n]{\left( \frac{n^2+1}{n^2+n+1} \right)^{n^2}} = \left( \frac{n^2+1}{n^2+n+1} \right)^{n} = \left[\left( 1-\frac{n}{n^2+n+1} \right)^{-\frac{n^2+n+1}{n}}\right]^{\frac{-n^2}{n^2+n+1}}\stackrel{n\to\infty}{\longrightarrow} e^{-1} < 1,$$
and therefore the series converges.
We used:

*

*$\left( 1-\frac{n}{n^2+n+1} \right)^{-\frac{n^2+n+1}{n}}\to e$, because $\frac{n}{n^2+n+1}\to 0,$

*$\frac{-n^2}{n^2+n+1}\to -1.$
