Question

# Determine whether the series converges or diverges and justify your answer. If the series converges, find its sum sum_{n=0}^inftyfrac{3^{n-2}}{4^{n+1}}

Series
Determine whether the series converges or diverges and justify your answer. If the series converges, find its sum
$$\sum_{n=0}^\infty\frac{3^{n-2}}{4^{n+1}}$$

2020-12-26
We have given a series,
$$\sum_{n=0}^\infty\frac{3^{n-2}}{4^{n+1}}$$
Let,
$$a_n=\frac{3^{n-2}}{4^{n+1}}$$
$$a_{n+1}=\frac{3^{n+1-2}}{4^{n+1+1}}$$
$$a_{n+1}=\frac{3^{n-1}}{4^{n+2}}$$
By the ratio test we know that,
$$\sum_{n=0}^\infty\frac{a_{n+1}}{a_n}=L$$
If L<1 then series converges, if L>1 then it diverges
So now we have,
$$\frac{a_{n+1}}{a_n}=\frac{\frac{3^{n-1}}{4^{n+2}}}{\frac{3^{n-2}}{4^{n+1}}}$$
$$\frac{a_{n+1}}{a_n}=\frac{3^{n-1}}{4^{n+2}}\cdot\frac{4^{n+1}}{3^{n-2}}$$
$$\frac{a_{n+1}}{a_n}=\frac{3^n3^{-1}}{4^n4^2}\cdot\frac{4^n4}{3^n3^{-2}}$$
$$\frac{a_{n+1}}{a_n}=\frac{1}{4}\cdot\frac{1}{3^{-1}}$$
$$\frac{a_{n+1}}{a_n}=\frac{3}{4}$$
$$\sum_{n=0}^\infty\frac{a_{n+1}}{a_n}=\frac{3}{4}$$
Hence, the series converges.