Trouble with Comparison test for Series

[Edder]

Hey everyone, I am having a bit of difficulty with a series question involving the comparison test.

\(\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{n}{(n+5)2^{n}}\)

The denominator should actually read (n+5) x 2^n. How would I go about comparing this. Would I use \(\displaystyle \frac{1}{n^2}\) or something else?
I appreciate any help, thanks.

 

[pka]

Hey everyone, I am having a bit of difficulty with a series question involving the comparison test.
\(\displaystyle \sum_{n=1}^{\infty} \frac{n}{(n+5)2^n}\)
How would I go about comparing this. Would I use \(\displaystyle \frac{1}{n^2}\) or something else?

I would use \(\displaystyle \dfrac{1}{2^n}\).

 

[soroban]

Hello, Edder!

I agree with pka.


\(\displaystyle \displaystyle\text{Since }\frac{n}{n+5} \:<\:1,\,\text{ then }\,\frac{n}{n+5}\!\cdot\!\frac{1}{2^n} \:<\:\frac{1}{2^n}\)

\(\displaystyle \displaystyle\text{Hence: }\:\sum^{\infty}_{n=1}\frac{n}{(n+5)2^n} \;<\;\sum^{\infty}_{n=1}\frac{1}{2^n}\)

 

[Edder]

Yeah that makes sense, thanks for the feedback.

 

[galactus]

This looks like a fun series to evaluate.

\(\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{nx^{n}}{n+5}\)

You could start with \(\displaystyle \displaystyle\sum_{n=1}^{\infty}x^{n}=\frac{x}{1-x}\), do some

differentiating, integrating, etc. hammer it into the required form, then let x=1/2.

Knowing that its sum will be less than \(\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{1}{2^{n}}=1\).