Question

Use the Limit Comparison Test to determine the convergence or divergence of the series. sum_{n=1}^inftyfrac{n}{(n+1)2^{n-1}}

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asked 2021-01-28
Use the Limit Comparison Test to determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\frac{n}{(n+1)2^{n-1}}\)

Answers (1)

2021-01-29

To determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\frac{n}{(n+1)2^{n-1}}\)
let
\(a_n=\frac{n}{(n+1)2^{n-1}}\)
\(b_n=\frac{1}{2^n}\)
By limit comparison test
let \(\sum a_n\) and \(\sum b_n\) be two series such that
\(\lim_{n\to\infty}\frac{a_n}{b_n}=c\quad0\)
then both series will converge or diverge together.
\(\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\frac{n}{(n+1)2^{n-1}}}{\frac{1}{2^n}}\)
\(=\lim_{n\to\infty}\frac{2n}{n+1}\)
\(=2\)
which is non zero and finite
Hence, both series converge or diverge together.
Since, \(\sum b_n\) is a geometric series with common ratio les than 1
Thus, it converges
Hence, by comparison test \(\sum a_n\) converges

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