[tex]\left|\dfrac{(-2)^n}{3^n+1}\right|=\dfrac{2^n}{3^n+1}<\dfrac{2^n}{3^n}=\left(\dfrac23\right)^n[/tex]
As [tex]n\to\infty[/tex], the sequence [tex]a_n=\left(\dfrac23\right)^n[/tex] converges to zero.
If you're talking about the infinite series
[tex]\displaystyle\sum_{n\ge0}\dfrac{(-2)^n}{3^n+1}[/tex]
well we've shown by comparison that this series must also converge because we know any geometric series [tex]\sum\limits_n r^n[/tex] will converge as long as [tex]|r|<1[/tex].
