<s>[INST] Convergence of $ \sum_{n=1}^{\infty} (\frac{n^2+1}{n^2+n+1})^{n^2}$ 
Find if the following series converge: $$\displaystyle \sum_{n=1}^{\infty} \left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}$$

What I did: 
$$a_n=\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}$$
$$b_n=\frac {1}{(n+1)^{n^2}}$$
$$\forall n\ge 1 : a_n < b_n \ \Rightarrow \ \displaystyle \sum_{n=1}^{\infty}\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}<\displaystyle \sum_{n=1}^{\infty}\frac {1}{(n+1)^{n^2}} \approx \displaystyle \sum_{n=1}^{\infty} \frac 1 {n^2}$$

$\displaystyle \sum_{n=1}^{\infty}b_n$ converges like $\displaystyle \sum_{n=1}^{\infty}\frac 1 {n^2}$ so $\displaystyle \sum_{n=1}^{\infty}a_n$ converges as well.

 [/INST] I would take the following approach:
$$\begin{align}\left (\frac{n^2+1}{n^2+n+1}\right )^{n^2} &= \left (1-\frac{n}{n^2+n+1}\right )^{n^2}\\ &=e^{n^2 \log{\left (1-\frac{n}{n^2+n+1}\right )}} \\ &\sim e^{-n^2 (1/n - 1/(2 n^2))}\\ &\sim e^{-n+1/2} \end{align}$$
Thus the series converges by comparison with a geometric series.
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