Evaluating a sum: $\sum_{n=0}^{\infty} nx^{2n+1}$

Question

My problem: I should calculate this sum: $\sum_{n=0}^{\infty} nx^{2n+1}$.

My solution: $\;$$ x \sum_{n=0}^{\infty} nx^{2n}$,

Substitution: $x^2 = y$

Then $\;$$ y^{1/2} \sum_{n=0}^{\infty} ny^{n}$, ...

Now I would focus just on $\;$$ \sum_{n=0}^{\infty} ny^{n} = y\sum_{n=0}^{\infty} ny^{n-1}$$, $

Because $\;$$ \sum_{n=0}^{\infty} y^{n} = 1/(1-y)$

EDIT

So $ \sum_{n=0}^{\infty} ny^{n-1} = 1/ (1-y)^2$

Finally $y^{1/2}\sum_{n=0}^{\infty} ny^{n-1} = y^{1/2}/ (1-y)^2 = x/(1-x^2)^2$

http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn — lab bhattacharjee, Mar 29 '17 at 09:43
Well, I just have problem with that $x^{2n}$, if I can use the substitution etc, so I don't consider it as a duplicate, it's different. — Leif, Mar 29 '17 at 09:46
@Leif you can write $x^{2n}=y^n$ where $y=x^2$ (as you did) so there is hardly a difference. — drhab, Mar 29 '17 at 09:53
Start by
$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$

By differentiation

$$\sum_{n=1}^\infty nx^{n-1} = \frac{1}{(1-x)^2}$$ — Zaid Alyafeai, Mar 29 '17 at 09:59

score 3 · Answer 1 · answered Mar 29 '17 at 10:05

By definition $\sum_{n=0}^\infty nx^{2n+1}=\lim_{N\to\infty} S_N$, where $S_N=\sum_{n=0}^N nx^{2n+1}$.

If $|x|>1$ the terms do not tend to zero and the sum does not converge.

Suppose $|x|<1$. Now $S_N-x^2S_N=\sum_{n=1}^{N}x^{2n+1}-Nx^{2N+3}=x\frac{1-x^{2N+2}}{1-x^2}-Nx^{2N+3}$.
(Since $x^{2n+1}$ appears once with coefficient $n$ and once with coefficient $-(n-1)$ if $n<N$.)

So we have $$S_N=x\frac{1-x^{2N+2}}{(1-x^2)^2}-\frac{Nx^{2N+3}}{1-x^2}.$$ The limit of this as $N\to\infty$ is $\frac x{(1-x^2)^2}$.

score 1 · Answer 2 · answered Mar 29 '17 at 10:09

1

Hint:

$$S-x^2S=(x^3+2x^5+3x^7+\cdots)-(x^5+2x^7+3x^9+\cdots)=x^3+x^5+x^7+\cdots$$

answered Mar 29 '17 at 10:09

The complete and rigorous development is made by @especiallylime. – Mar 29 '17 at 10:10

Evaluating a sum: $\sum_{n=0}^{\infty} nx^{2n+1}$

2 Answers2