# Thread: representation of a power series

1. ## representation of a power series

express the following as a power series:

$\displaystyle \int \frac{ln(1-t)}{t} dt$

so far I have

$\displaystyle \mbox{let} \ g(t) = ln(1-t) \rightarrow g(t) = \int g'(t) dt = \int -\frac{1}{1-t} \ dt= \int - \sum_{n=0}^{\infty} t^n \ dt$

$\displaystyle - \sum_{n=0}^{\infty} \frac{t^{n-1}}{n} = - \sum_{n=1}^{\infty} \frac{t^{n}}{n}$

now putting it back into the original:

$\displaystyle \int - \sum_{n=1}^{\infty} \frac{t^n}{n} \cdot \frac{1}{t} \ dt = \int -\frac{1}{n} \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt$

$\displaystyle -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \sum_{n=1}^{\infty} \frac{t^{n-2}}{n-1} = -\sum_{n=1}^{\infty} \frac{t^{n-2}}{n^2-n}$

is this correct?

2. Originally Posted by lllll
express the following as a power series:

$\displaystyle \int \frac{ln(1-t)}{t} dt$

so far I have

$\displaystyle \mbox{let} \ g(t) = ln(1-t) \rightarrow g(t) = \int g'(t) dt = \int -\frac{1}{1-t} \ dt= \int - \sum_{n=0}^{\infty} t^n \ dt$

$\displaystyle - \sum_{n=0}^{\infty} \frac{t^{n-1}}{n} = - \sum_{n=1}^{\infty} \frac{t^{n}}{n}$

now putting it back into the original:

$\displaystyle \int - \sum_{n=1}^{\infty} \frac{t^n}{n} \cdot \frac{1}{t} \ dt = \int -\frac{1}{n} \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt$

$\displaystyle -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \sum_{n=1}^{\infty} \frac{t^{n-2}}{n-1} = -\sum_{n=1}^{\infty} \frac{t^{n-2}}{n^2-n}$

is this correct?
Yes as far as I can see it is correct good job

3. ## Wait

Originally Posted by lllll
express the following as a power series:

$\displaystyle \int \frac{ln(1-t)}{t} dt$

so far I have

$\displaystyle \mbox{let} \ g(t) = ln(1-t) \rightarrow g(t) = \int g'(t) dt = \int -\frac{1}{1-t} \ dt= \int - \sum_{n=0}^{\infty} t^n \ dt$

$\displaystyle - \sum_{n=0}^{\infty} \frac{t^{n-1}}{n} = - \sum_{n=1}^{\infty} \frac{t^{n}}{n}$

now putting it back into the original:

$\displaystyle \int - \sum_{n=1}^{\infty} \frac{t^n}{n} \cdot \frac{1}{t} \ dt = \int -\frac{1}{n} \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt$

$\displaystyle -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \sum_{n=1}^{\infty} \frac{t^{n-2}}{n-1} = -\sum_{n=1}^{\infty} \frac{t^{n-2}}{n^2-n}$

is this correct?
I do see one problem $\displaystyle \int{t^{n-1}}dt=\frac{t^n}{n}$

4. Originally Posted by lllll
$\displaystyle - \sum_{n=0}^{\infty} \frac{t^{n-1}}{n} = - \sum_{n=1}^{\infty} \frac{t^{n}}{n}$
Careful here.

The thing is, you should have memorized the power series representation for $\displaystyle \ln(1-x),$ that is $\displaystyle -\sum_{k\,=\,1}^\infty\frac{x^k}k$ for $\displaystyle |x|<1.$ From here you can easily find $\displaystyle \frac{\ln(1-x)}x$ as a series, then interchange sum by integral and you're done.

There's no much here, simple calculations, it's easy.