Prove: $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2} < \infty$ without L'Hôpital's Given $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2}$, prove that it converges. 
I tried to use the Ratio test. 
I got a terrible algebraic expression: $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \left(\frac{n^4 +2n^3 +3n^2 + 2n +2}{n^4 + 4n^3 + 7n^2 + 6n +3}\right)\cdot \left(\frac{n^2 + 2n + 2}{n^2 + 3n + 3}\right)^{2n +1}$$
Now problem is I'm not allowed to use L'Hoptial and I don't really know what to do with this disaster.  

By Taylor's expansion we have
$$\left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2}=\left(1-\frac{n}{n^{2} +n +1}\right)^{n^2}=e^{n^2 \log\left(1-\frac{n}{n^{2} +n +1}\right)}=e^{n^2 \left(\frac{-n}{n^{2} +n +1}+O(1/n^2)\right)}\sim \frac c{e^n}$$
and then refer to limit comparison test.
