0
$\begingroup$

I have calculated that $\sum^\infty_1 nx^n$ converges for $|x| < 1$ using ratio test etc. I have also determined that the sum can be expressed simply in the form $x/(1-x)^2$ in this range, although I got this formula mainly through observation so my first question is if there is a more concrete way of determining this using the partial sums of the sequence or something?

Now, I would like to calculate $\sum^\infty_1 nx^{n-1}$ for $|x|<1$ also, and I am not sure how to go about doing this, which I think is because I have not derived my expression above through a proper method yet.

Any hints or advice are much appreciated, thanks!

  • 1
    $$\frac{\mathsf d}{\mathsf dx} \left[\sum_{n=0}^\infty x^n\right] =\quad ? $$ – 2017-01-13

1 Answers 1

1

For all $x\in\mathbb{R}-\{1\}$ and $n\in\mathbb{N^\star}$, we have :

$$\sum_{k=0}^{n-1}x^k=\frac{1-x^n}{1-x}$$

Derivation with respect to $x$ gives us (for $n\ge2$) :

$$\sum_{k=1}^{n-1}kx^{k-1}=\frac{-nx^{n-1}(1-x)+(1-x^n)}{(1-x)^2}=\frac{(n-1)x^n-nx^{n-1}+1}{(1-x)^2}$$

Now, if we make the assumption that $\vert x\vert<1$, we know that $\lim_{n\to\infty}nx^n=0$ and therefore :

$$\sum_{k=1}^\infty kx^{k-1}=\frac{1}{(1-x)^2}$$

and of course (multiplicating by $x$) :

$$\sum_{k=1}^\infty kx^k=\frac{x}{(1-x)^2}$$