Solve $\sum_{i=1}^n n*x^{2n}$

Question

I know that $\sum_{i=0}^n n*x^n = \frac{x}{(1-x)^2}$ but the factor of 2 in the exponential makes it a riemann zeta function? Can anyone shed some light on this?

Answer 1

Hint: Replace $x$ with $x^2$.

Answer 2

Note that, for a geometric series with ratio $r$,

$$\sum_{n=0}^\infty r^n = \frac{1}{1-r}$$

and the derivative is given by

$$\sum_{n=0}^\infty nr^{n-1} = \frac{1}{(1-r)^2}$$

and multiplication by $r$ gives

$$\sum_{n=0}^\infty nr^{n} = \frac{r}{(1-r)^2}$$

Let $r=x^2$. Then we have

$$\sum_{n=0}^\infty nx^{2n} = \frac{x^2}{(1-x^2)^2}$$

(Equivalently we can start the summation at $n=1$ since the summand for $n=0$ is $0$. Thus, we have your desired expression.)

---

Note that this is not the Riemann zeta function since it is defined by

$$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$$

It's a rather different function; it's moreso a generalization of the idea behind harmonic series than it is of geometric series.

Answer 3

First of all the Riemann zeta function is $$\zeta(s)=\sum_{n=1}^{\infty}\frac1{n^s}$$ And your sum is $$f(x)=\sum_{i=1}^{n}ix^{2i}$$ You already know that $$\sum_{i=1}^nix^i=\frac{x}{(1-x)^2}$$ So just replace $x$ with $x^2$ to get $$f(x)=\frac{x^2}{(1-x^2)^2}$$