Proving that the following series is convergent: $\sum\limits_{i=1}^\infty \left({n^2+1\over n^2+n+1}\right)^{n^2}$

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.

The $n$th term is $\left(1-\dfrac{n}{n^2+n+1}\right)^{n^2}<\left(\left(1-\dfrac{1}{2n}\right)^n\right)^n$. I would suggest a limit comparison test with $1/e^{n/2}$.

$$ \left(1-\dfrac{n}{n^2+n+1}\right)^{n^2}\lt\exp\left(-\dfrac{n}{n^2+n+1}\cdot n^2\right)\lt\mathrm e^{-n+1} $$