1. ## 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

2. 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?

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

4. 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 $\bigg|\frac{x-a}{1-a}\bigg|<1$

5. 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 $x < 1$

6. 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?

7. 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 $a \ne 1$ & $x \ne a$