[tex]\displaystyle\sum_{n\ge0}\frac{2e^n}{e^{2n}+1}=\sum_{n\ge0}\frac2{e^n+\frac1{e^n}}[/tex]
As [tex]n\to\infty[/tex], [tex]\dfrac1{e^n}[/tex] becomes insignificant, so you can compare this series to
[tex]\displaystyle\sum_{n\ge0}\frac2{e^n}[/tex]
since
[tex]e^n<e^n+\dfrac1{e^n}\implies\dfrac2{e^n}>\dfrac2{e^n+\frac1{e^n}}[/tex]
This comparison series converges as it's geometric:
[tex]\displaystyle\sum_{n\ge0}\frac2{e^n}=2\sum_{n\ge0}\left(\frac1e\right)^n=\frac{2e}{e-1}[/tex]
Since this convergent series is smaller than the first series, the first series must also be convergent.
