Answer:
Limit = 2
Step-by-step explanation:
[tex]\sum _{n=1}^{\infty }\frac{2^n}{3^n}\\= \sum _{n=1}^{\infty \:}\left(\frac{2}{3}\right)^n\\\\= \sum _{n=0}^{\infty \:}\left(\frac{2}{3}\right)^n-\left(\frac{2}{3}\right)^0[/tex]
[tex]\sum _{n=0}^{\infty \:}\left(\frac{2}{3}\right)^n=3,\\\\= 3-\left(\frac{2}{3}\right)^0\\= 2[/tex]
In this case it does converge, as 2/3 is between 1 and 0
