Answered: "Determine whether the series converges or diverges and..."

permaneceerc

Answered question

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}}

Answer & Explanation

Usamah Prosser

Skilled2020-12-26Added 86 answers

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,
$\frac{a_{n + 1}}{a_{n}} = \frac{\frac{3^{n - 1}}{4^{n + 2}}}{\frac{3^{n - 2}}{4^{n + 1}}}$
$\frac{a_{n + 1}}{a_{n}} = \frac{3^{n - 1}}{4^{n + 2}} \cdot \frac{4^{n + 1}}{3^{n - 2}}$
$\frac{a_{n + 1}}{a_{n}} = \frac{3^{n} 3^{- 1}}{4^{n} 4^{2}} \cdot \frac{4^{n} 4}{3^{n} 3^{- 2}}$
$\frac{a_{n + 1}}{a_{n}} = \frac{1}{4} \cdot \frac{1}{3^{- 1}}$
$\frac{a_{n + 1}}{a_{n}} = \frac{3}{4}$
$\sum_{n = 0}^{\infty} \frac{a_{n + 1}}{a_{n}} = \frac{3}{4}$
Hence, the series converges.

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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