# Thread: How to show this sequence converges

1. ## How to show this sequence converges

Prove:

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

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?

2. Hint :differentiate both sides of the identity

$1+x+x^2+...x^n=\frac{x^{n+1}-1}{x-1}$

3. I can't use differentiation. :-(

4. Originally Posted by paupsers
Prove:

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

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?
$S=1+2x+3x^2+...+nx^{n-1}+...\implies \frac{S-1}{x}=2+3x+4x^2+...+nx^{x-2}+...$

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

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

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

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

Obviously, this method is not as fast as differentiation.

5. Look this is simple.
If $\left| x \right| < 1\, \Rightarrow \,f(x) = \sum\limits_{k = 0}^\infty {x^k } = \frac{1}{{1 - x}}$
then $f'(x) = \sum\limits_{k = 0}^\infty {kx^{k - 1} } = \frac{1}{{\left( {1 - x} \right)^2 }}$

6. Originally Posted by Plato
Look this is simple.
If $\left| x \right| < 1\, \Rightarrow \,f(x) = \sum\limits_{k = 0}^\infty {x^k } = \frac{1}{{1 - x}}$
then $f'(x) = \sum\limits_{k = 0}^\infty {kx^{k - 1} } = \frac{1}{{\left( {1 - x} \right)^2 }}$
That works well, but the only problem is:

Originally Posted by paupsers
I can't use differentiation. :-(

7. 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?