# Taylor serie

• June 23rd 2008, 06:56 AM
Klaus
Taylor serie
Hello,
How to show that the Taylor serie of $\frac{1}{1-x}$ converges ?
I found that the serie was $\sum_{n=1}^{\infty} \frac{(x-a)^{n-1}}{(1-a)^n}$.

Should I use Lagrange's remainder ?

Thank you :)
• June 23rd 2008, 07:02 AM
Mathstud28
Quote:

Originally Posted by Klaus
Hello,
How to show that the Taylor serie of $\frac{1}{1-x}$ converges ?
I found that the serie was $\sum_{n=1}^{\infty} \frac{(x-a)^{n-1}}{(1-a)^n}$.

Should I use Lagrange's remainder ?

Thank you :)

$\frac{1}{a-x}=\frac{1}{a}\frac{1}{1-\frac{x}{a}}=\frac{1}{a}\sum_{n=0}^{\infty}\left(\ frac{x}{a}\right)^n\quad\forall{x,a}\backepsilon\t ext{ }\left|\frac{x}{a}\right|<1$

This is just an extension of the geometric series.

Is this what you were asking?
• June 23rd 2008, 08:43 AM
Klaus
In fact I would like to show that
$\sum_{n=1}^{\infty} \frac{(x-a)^{n-1}}{(1-a)^n}$
converges...
• June 23rd 2008, 08:47 AM
Mathstud28
Quote:

Originally Posted by Klaus
In fact I would like to show that
$\sum_{n=1}^{\infty} \frac{(x-a)^{n-1}}{(1-a)^n}$
converges...

Ok, except it doesnt always now does it?

If $a=0$ and $x=10$ for example

But utilizing the Root test we can see that

$\sum_{n=1}^{\infty}\frac{(x-a)^{n-1}}{(1-a)^n}$ converges

iff $\bigg|\frac{x-a}{1-a}\bigg|<1$
• June 23rd 2008, 08:48 AM
colby2152
Quote:

Originally Posted by Klaus
In fact I would like to show that
$\sum_{n=1}^{\infty} \frac{(x-a)^{n-1}}{(1-a)^n}$
converges...

$\sum_{n=1}^{\infty} \frac{(x-a)^{n-1}}{(1-a)^n}$

simplifies to

$\sum_{n=1}^{\infty} \frac{(x-a)^{n}}{(x-a)(1-a)^n}$

$\frac{1}{x-a} \sum_{n=1}^{\infty} \left(\frac{(x-a)}{(1-a)}\right)^n$

It converges as a geometric sum if $|x - a| < |1 - a|$ which means it will definitely converge for values of $x < 1$
• June 23rd 2008, 08:50 AM
Mathstud28
Quote:

Originally Posted by colby2152
$\sum_{n=1}^{\infty} \frac{(x-a)^{n-1}}{(1-a)^n}$

simplifies to

$\sum_{n=1}^{\infty} \frac{(x-a)^{n}}{(x-a)(1-a)^n}$

$\frac{1}{x-a} \sum_{n=1}^{\infty} \left(\frac{(x-a)}{(1-a)}\right)^n$

It converges as a geometric sum if $x - a < 1 - a$ which means it will definitely converge for values of $x < 1$

Didn't you neglect an absolute value there?
• June 23rd 2008, 09:20 AM
colby2152
Quote:

Originally Posted by Mathstud28
Didn't you neglect an absolute value there?

Good point or else there could be an infinite large negative sum.

Also, note that $a \ne 1$ & $x \ne a$