Series

[Brain0991]

Determine the convergence or divergence of the series, and if it converges find the sum.

its from n=1 to infinity

[1/(5[sup:m8bbvxog]n[/sup:m8bbvxog])] - [1/(n(n+1))]

I do not know how to start this, I guess it could be a telescoping series but I am not certain.

Should I start by getting a common denominator and subtracting?

 

[BigGlenntheHeavy]

\(\displaystyle You \ have \ a \ geometric \ series \ and \ a \ telescoping \ series, \ hence:\)

\(\displaystyle \sum_{n=1}^{\infty}ar^n \ = \ \frac{a}{1-r}, \ 0 \ < \ |r| \ < \ 1.\)

\(\displaystyle Note: \ \frac{1}{5^n} \ = \ \bigg(\frac{1}{5}\bigg)^n\)

\(\displaystyle Now, \ altogether, \ we \ have \ \sum_{n=1}^{\infty}\bigg[\frac{1}{5^n}-\frac{1}{n(n+1)}\bigg] \ = \ \sum_{n=1}^{\infty}\bigg(\frac{1}{5}\bigg)^n \ - \ \sum_{n=1}^{\infty}\bigg[\frac{1}{n}-\frac{1}{n+1}\bigg] \ = \ \frac{1}{4} \ - \ 1 \ = \ -\frac{3}{4}.\)

 

[Brain0991]

Thank you so much. I would have never saw that geometric series. I will look out for it from now on.