Prove:

The sequence $\displaystyle b_n=1+2x+3x^2+...+nx^{n-1}, |x|<1$ converges to $\displaystyle (1-x)^{-2}$

I don't even know what method to use to show:

1. that this converges, and

2. that it converges to that value.

Any help with this one?

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- Oct 30th 2009, 03:07 PMpaupsersHow to show this sequence converges
Prove:

The sequence $\displaystyle b_n=1+2x+3x^2+...+nx^{n-1}, |x|<1$ converges to $\displaystyle (1-x)^{-2}$

I don't even know what method to use to show:

1. that this converges, and

2. that it converges to that value.

Any help with this one? - Oct 30th 2009, 03:21 PMBruno J.
Hint :differentiate both sides of the identity

$\displaystyle 1+x+x^2+...x^n=\frac{x^{n+1}-1}{x-1}$ - Oct 30th 2009, 03:22 PMpaupsers
I can't use differentiation. :-(

- Oct 30th 2009, 03:36 PMredsoxfan325
$\displaystyle S=1+2x+3x^2+...+nx^{n-1}+...\implies \frac{S-1}{x}=2+3x+4x^2+...+nx^{x-2}+...$

Now subtract $\displaystyle 1+x+x^2+...+x^{n-2}+...=\frac{1}{1-x}$ from both sides:

$\displaystyle \frac{S-1}{x}-\frac{1}{1-x}=1+2x+3x^2+...+(n-1)x^{n-2}+...$

But the right side of the equation now equals $\displaystyle S$, so solve for it in the equation below.

$\displaystyle \frac{S-1}{x}-\frac{1}{1-x}=S$

Obviously, this method is not as fast as differentiation. - Oct 30th 2009, 03:43 PMPlato
Look this is simple.

If $\displaystyle \left| x \right| < 1\, \Rightarrow \,f(x) = \sum\limits_{k = 0}^\infty {x^k } = \frac{1}{{1 - x}}$

then $\displaystyle f'(x) = \sum\limits_{k = 0}^\infty {kx^{k - 1} } = \frac{1}{{\left( {1 - x} \right)^2 }}$ - Oct 30th 2009, 03:45 PMredsoxfan325
- Nov 2nd 2009, 11:44 AMpaupsers
What's a good method to show that the limit exists for this sequence? I've tried showing that it's contractive, but no luck there. Tried the ratio test, didn't work for me. Any ideas?