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

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

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Malena
Answered 2021-01-29 Author has 26120 answers

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

Not exactly what you’re looking for?
Ask My Question
39
 
content_user
Answered 2021-12-27 Author has 9769 answers

Answer is given below (on video)

0

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more
...