$\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?

$$a_n=\left(\frac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2}\implies \log(a_n)=n^2\log\left(\frac{ n^2 + 1}{n^2 +n + 1}\right)$$
For large $n$
$$\log(a_n)=-n+\frac{1}{2}+\frac{2}{3 n}-\frac{3}{4 n^2}-\frac{1}{5
   n^3}+O\left(\frac{1}{n^4}\right)$$ Repeat it for $a_{n+1}$ to obtain
$$\log(a_{n+1})-\log(a_n)=-1-\frac{2}{3 n^2}+\frac{13}{6 n^3}+O\left(\frac{1}{n^4}\right)$$
$$\frac{a_{n+1}}{a_{n}}=e^{\log(a_{n+1})-\log(a_n)}=\frac 1e \left(1-\frac{2}{3 n^2}+\frac{13}{6 n^3} \right)+O\left(\frac{1}{n^4}\right)$$ So, it converges.
Now, if you make
$$\sum_{n=0}^\infty a_n=\sum_{n=0}^p a_n+\sum_{n=p+1}^\infty a_n$$
$$\sum_{n=p+1}^\infty a_n\sim \sum_{n=p+1}^\infty e^{\frac{1}{2}-n}=\frac{e^{\frac{1}{2}-p}}{e-1}$$
If you make $p=4$, the approximation gives
$$\frac{31179700687749571657625354575093}{15170464778704564632455243642733}+\frac{1}{(e-
   1) e^{7/2}}\sim 2.072864$$ while the "excat" result is $2.074636$.
