<s>[INST] 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.  
 [/INST] 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 $e^{-1}<1$.
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