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}}

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asked 2020-12-25
Determine whether the series converges or diverges and justify your answer. If the series converges, find its sum
\(\sum_{n=0}^\infty\frac{3^{n-2}}{4^{n+1}}\)

Answers (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.
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