# Thread: Power series, term-by-term integration

1. ## Power series, term-by-term integration

Evaluate

∑ 1 / [n(n+1) $5^n$]
n=1
===================================

Attempt:

$x^n$ = 1/(1-x) for |x|<1
n=0

Take the indefinite integral of both sides...

$x^{n+1}$ /(n+1) + C= -ln(1-x)
n=0
Put x=0 => C=0

$x^{n+1}$ /(n+1) = -ln(1-x)
n=0

Divide both sides by x^2

$x^{n-1}$ /(n+1) = - ln(1-x) / $x^2$
n=0

$x^{n-1}$ /(n+1) = - [ln(1-x) / $x^2$] - 1/x
n=1

How to continue?

Thanks for helping!

2. Originally Posted by kingwinner
Evaluate

∑ 1 / [n(n+1) $5^n$]
n=1
===================================

Attempt:

$x^n$ = 1/(1-x) for |x|<1
n=0

Take the indefinite integral of both sides...

$x^{n+1}$ /(n+1) + C= -ln(1-x)
n=0
Put x=0 => C=0

$x^{n+1}$ /(n+1) = -ln(1-x)
n=0

Divide both sides by x^2

$x^{n-1}$ /(n+1) = - ln(1-x) / $x^2$
n=0

$x^{n-1}$ /(n+1) = - [ln(1-x) / $x^2$] - 1/x
n=1

How to continue?

Thanks for helping!

I'd begin with $\sum^\infty_{n=1}x^{n-1}=\frac{1}{1-x},\,|x|<1\Longrightarrow \sum^\infty_{n=1}\frac{x^n}{n}=-\ln(1-x),\,|x|<1$ $\Longrightarrow \sum^\infty_{n=1}\frac{x^{n+1}}{n(n+1)}=(1-x)\ln(1-x)+x-1,\,|x|<1$ , and now just substitute $x=\frac{1}{5}$ ...

Tonio

3. Originally Posted by tonio
I'd begin with $\sum^\infty_{n=1}x^{n-1}=\frac{1}{1-x},\,|x|<1\Longrightarrow \sum^\infty_{n=1}\frac{x^n}{n}=-\ln(1-x),\,|x|<1$ $\Longrightarrow \sum^\infty_{n=1}\frac{x^{n+1}}{n(n+1)}=(1-x)\ln(1-x)+x-1,\,|x|<1$
How did you get the -1 at the end? Are you sure it is -1, instead of 0?

, and now just substitute $x=\frac{1}{5}$ ...
But if we just substitute x=1/5, we will get $1/5^{n+1}$ which is not what we want. We need $1/5^n$

Thanks!

4. Originally Posted by kingwinner
How did you get the -1 at the end? Are you sure it is -1, instead of 0?

A mistake, indeed. Erase it.

But if we just substitute x=1/5, we will get $1/5^{n+1}$ which is not what we want. We need $1/5^n$

This is not problem at all: $\frac{1}{5^{n+1}}=\frac{1}{5}\cdot\frac{1}{5^n}$ ...

Tonio

Thanks!
.