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.

 A: 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.
A: This is likely what I'd have a Calculus student do. Use the root test:
$$\lim_{n \rightarrow \infty} \left| \left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}\right|^\frac{1}{n} = \lim_{n \rightarrow \infty} \left( \frac{n^2+1}{n^2+n+1}\right)^n$$
$$\begin{array} \\
\lim_{n \rightarrow \infty} \ln \left(\left( \frac{n^2+1}{n^2+n+1}\right)^n\right) &= \lim_{n \rightarrow \infty} n \ln \left( \frac{n^2+1}{n^2+n+1}\right) \\
&= \lim_{n \rightarrow \infty} \frac{\ln \left( \frac{n^2+1}{n^2+n+1}\right)}{\frac{1}{n}} \\
&= \lim_{n \rightarrow \infty} \frac{\left( \frac{n^2+n+1}{n^2+1}\right)\left(\frac{(n^2+n+1)(2n)-(n^2+1)(2n+1)}{(n^2+n+1)^2}\right)}{-\frac{1}{n^2}} \\
&= \lim_{n \rightarrow \infty} -\frac{n^2(2n^3+2n^2+2n - 2n^3 - n^2 - 2n - 1)}{(n^2+1)(n^2+n+1)} \\
&= \lim_{n \rightarrow \infty} -\frac{n^2(n^2-1)}{(n^2+1)(n^2+n+1)} = -1
\end{array}$$
Now since the limit of $\ln (f(n)) = -1$, we have that the limit of $f(n) = \frac{1}{e} < 1$, and the series is absolutely convergent.
