If $\sum_{k=1}^{\infty}kx^k$ = 30, compute x.

Question

If $\sum_{k=1}^{\infty}kx^k$ = 30, compute x.
I don't think I can find $x$ because using $a/(1-r)$ is only one equation for two variables.
Looking it up on wolfram alpha, it says $x/(x-1)^2$ when $|x| < 1$. Can someone explain to me please?

Answer 1

You can observe what happens to the summation when you expand the first terms:

$$ \sum^{\infty}_{k=1} kx^k = 1(x) + 2(x^2) + 3(x^3) + \dots $$

This can be rearranged in a way that will yield in a sum of geometric series: $$ 1(x) + 2(x^2) + 3(x^3) ... = x + (x^2 + x^2) + (x^3 + x^3 + x^3) + \dots $$

Now observe what taking the first element of each term, you have a geometric series that starts a $x^1$.

But starting from the second term, taking the second element from all terms you also have a geometric series that starts on $x^2$.

So you actually have a sum of series, each one starting at plus one higher coefficient.

$$ \sum^{\infty}_{k=1} kx^k = \sum^{\infty}_{k=1}\sum^{\infty}_{j=k} x^j $$

Now you can solve the first summation: $a_0 = x^k$, then: $$ \sum^{\infty}_{j=k} x^j = \frac{x^k}{1-x} $$

Now solve the second summation:

$$ \sum^{\infty}_{k=1}\frac{x^k}{1-x} = \frac{1}{1-x}\sum^{\infty}_{k=1}{x^k} = \frac{x}{(1-x)^2} $$

Answer 2

Hint

$$\sum_{k=1}^{\infty}kx^k=x\sum_{k=1}^{\infty}kx^{k-1}=x\left(\sum_{k=1}^{\infty}x^{k}\right)'$$

Answer 3

$$\sum_{k=1}^{\infty}kx^k = 30 \implies \frac{x}{(x - 1)^2} = 30\quad\text{ for}\quad |x|<1\quad\implies\quad x=\frac{5}{6}$$