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.  
 A: Try the root criterion.
Prove that
$$\sqrt[n]{\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}}=\left(1-\frac{n}{n^2+n+1}\right)^n$$
and that this tends to $L<1$ as $n\to \infty$. Then, the criterion implies that the sum converges.
To find the limit, remember that if $a_n\to 0$ then
$$(1+a_n)^{\frac1{a_n}}\to e.$$
So
$$\left(1-\frac{n}{n^2+n+1}\right)^n=\left(1-\frac{n}{n^2+n+1}\right)^{\left(-\frac{n^2+n+1}{n}\right)\cdot \left(-\frac{n}{n^2+n+1}\right)\cdot n}=$$
$$=\left[\left(1-\frac{n}{n^2+n+1}\right)^{-\frac{n^2+n+1}{n}}\right]^{ \left(-\frac{n}{n^2+n+1}\right)\cdot n}.$$
But the bracketed expression is exactly of the form
$$(1+a_n)^{\frac1{a_n}},\quad a_n\to 0,$$
so it tends to $e$. And the expression in the exponent clearly goes to $-1$, so the limit is $e^{-1}<1$.
A: Another approach using the Root test.
Let $L = \lim_{n\to\infty} \sqrt[n]{\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}} $
$$ \begin{align}
L &= \lim_{n\to\infty} \exp \log \sqrt[n]{\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}} \\
&= \exp \lim_{n\to\infty}  \frac{\log \left(\left( \frac{n^2+1}{n^2+n+1} \right)^{n^2}\right)}{n} \\
&= \exp \lim_{n\to\infty}  \frac{n^2 \log \frac{n^2+1}{n^2+n+1}}{n} \\
&= \exp \lim_{n\to\infty} n \log \frac{n^2+1}{n^2+n+1}  \tag{1} \\
&= \exp(-1) \\
\end{align}$$
and since $L = \exp(-1) < 1$  the series is absolutely convergent by the root test.
* (1) can be shown by the fact that the Laurent expansion for $n\log\frac{n^2+1}{n^2+n+1}$ at $n=\infty$ is $-1 + \textrm{O}(\frac{1}{n})$.
A: 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.
