Is there a conclusive convergence test for $\sum_{n=1}^\infty (e^{2/n^2}-1)$?

Question

I have tried testing for convergence using the common methods but none of them were conclusive. Wolfram Alpha tells me $$\sum_{n=1}^\infty \left (e^{2/n^2}-1\right )$$ does approximately converge to 7.8597 but it gives no further explanation as to why or how. Is there a trick or some standard series that makes this solvable?

Thanks in advance!

Answer 1

Since $$ \lim_{h\to0}\frac{e^{2h}-1}h=2, $$ you have $$ \lim_{n\to\infty}\frac{e^{2/n^2}-1}{1/n^2}=2. $$ Therefore, since the series $\sum_{n=1}^\infty\frac1{n^2}$ converges, the series $\sum_{n=1}^\infty\left(e^{2/n^2}-1\right)$ is also convergent.

Answer 2

When $n\to +\infty$, we have: $$e^{\frac{2}{n^2}}-1\,\,\sim\,\,\frac{2}{n^2}$$ as the asympotic $e^x-1\,\,\sim\,\, x$ when $x\to 0$ holds.

So, the series $\sum_{n=1}^{+\infty}\left(e^{\frac{2}{n^2}}-1\right)$ converge by the asynptotic test.