<s>[INST] Proving that the following series is convergent: $\sum\limits_{i=1}^\infty \left({n^2+1\over n^2+n+1}\right)^{n^2}$ Can someone please help me prove that this series is convergent?
$$ \sum_{i=1}^\infty \left({n^2+1\over n^2+n+1}\right)^{n^2} $$
I guess I'm supposed to show that the limit of the sequence is an "e" limit, that means something of the kind: $$ \left( 1\pm{1 \over a} \right)^a $$
But how? I came to this state by now and that's where I'm stuck:
$$ \left( {n^2+n+1-n \over {n^2+n+1}} \right)^{n^2} = \left( 1+{n \over {n^2+n+1}} \right)^{n^2} $$
 [/INST] Note that $${\left( \dfrac{n^2+n+1}{n^2+1}\right)^{n^2}}={\left(1+ \dfrac{n}{n^2+1}\right)^{\frac{n^2+1}{n}\cdot\frac{n^3}{n^2+1}}}\geqslant 2^{\frac{n}{2}},$$ because $2<\left(1+ \dfrac{n}{n^2+1}\right)^{\frac{n^2+1}{n}}<3$ and $\frac{n^3}{n^2+1}\geqslant \frac{n}{2}$ for $n\geqslant 1.$ Therefore,
$$\left( \dfrac{n^2+1}{n^2+n+1}\right)^{n^2}\leqslant {\frac{1}{2^{\frac{n}{2}}}}=\left(\frac{1}{\sqrt{2}}\right)^n$$
and the series
$$\sum_{n=1}^\infty \left({n^2+1\over n^2+n+1}\right)^{n^2}$$
converges by comparison test.
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