Find the sum $\sum_{n=1}^\infty \frac{n}{(1+x)^{2n+1}}$

Question

Find the sum $$\sum_{n=1}^\infty \frac{n}{(1+x)^{2n+1}}.$$

Indicating the interval of convergence for $x$.

My attempt:

Let $ t=\frac{1}{x+1}$. Then, applying the root test,

$$\lim_{n\to \infty} \{n t^{2n+1}\}^{1/n} = |t|^2 < 1 \iff |t| <1.$$

Then, we have that $|x-1| >1 \iff x < -2 \text{ or } x >0$.

Now, consider the series

$$\sum_{n=1}^\infty n t^n = \frac{t}{(t-1)^2}, \quad |t|<1.$$

So, since $|t|< 1 \Rightarrow |t^2| <1$, $$\sum_{n=1}^\infty n t^{2n }= \frac{t^2}{(t^2-1)^2}, \quad |t|<1.$$

Multiplying by $t,$

$$\sum_{n=1}^\infty n t^{2n +1 }= \frac{t^3}{(t^2-1)^2}, \quad |t|<1.$$

If we substitute back, we have what we want.

I want to know if my steps are correct. I have doubts about the interval of convergence part.

Thanks for your effort!

Answer 1

Starting with the series \begin{align} \sum_{n=1}^{\infty} n \, t^{n} = \frac{t}{(1-t)^{2}} \end{align} then it is seen that \begin{align} \sum_{n=1}^{\infty} \frac{n}{(1+x)^{2n+1}} = \frac{1}{(1+x)^{3}} \cdot \frac{(1+x)^{4}}{[(1+x)^{2}-1]^{2}} = \frac{1+x}{x^{2} \, (2+x)^{2}}. \end{align} The series does not converge for $x \in\{0, -2\}$.

Answer 2

First recall the basic formula for the sum of a geometric series:

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

Differentiating, we obtain:

$$\sum_{n=1}^{\infty}nz^{n-1}=\frac{1}{(1-z)^2}.$$

Multiplying both sides by $z$ yields:

$$\sum_{n=1}^{\infty}nz^{n}=\frac{z}{(1-z)^2}.$$

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Now, we can rewrite the series $S(x)=\sum_{n=1}^{\infty}\frac{n}{(1+x)^{2n+1}}$ as a finite sum of series that are summable via the formulas given above s follows:

$$\begin{align} S(x) &=\sum_{n=1}^{\infty}\frac{n}{(1+x)^{2n+1}}\\ &=\frac12\sum_{n=1}^{\infty}\frac{2n}{(1+x)^{2n+1}}\\ &=\frac12\sum_{n=1}^{\infty}\frac{2n+1-1}{(1+x)^{2n+1}}\\ &=\frac12\sum_{n=1}^{\infty}\frac{2n+1}{(1+x)^{2n+1}}-\frac12\sum_{n=1}^{\infty}\frac{1}{(1+x)^{2n+1}}\\ &=\frac12\sum_{n=1}^{\infty}\frac{2n}{(1+x)^{2n}}+\frac12\sum_{n=1}^{\infty}\frac{2n+1}{(1+x)^{2n+1}}-\frac12\sum_{n=1}^{\infty}\frac{1}{(1+x)^{2n}}\\ &~~~~~ -\frac12\sum_{n=1}^{\infty}\frac{1}{(1+x)^{2n+1}}-\frac12\sum_{n=1}^{\infty}\frac{2n}{(1+x)^{2n}}+\frac12\sum_{n=1}^{\infty}\frac{1}{(1+x)^{2n}}\\ &=\frac12\sum_{n=2}^{\infty}\frac{n}{(1+x)^{n}}-\frac12\sum_{n=2}^{\infty}\frac{1}{(1+x)^{n}}-\sum_{n=1}^{\infty}\frac{n}{(1+x)^{2n}}+\frac12\sum_{n=1}^{\infty}\frac{1}{(1+x)^{2n}}. \end{align}$$

In the last line above, let $z=\frac{1}{1+x}$:

$$\begin{align} S(x) &=\frac12\sum_{n=2}^{\infty}\frac{n}{(1+x)^{n}}-\frac12\sum_{n=2}^{\infty}\frac{1}{(1+x)^{n}}-\sum_{n=1}^{\infty}\frac{n}{(1+x)^{2n}}+\frac12\sum_{n=1}^{\infty}\frac{1}{(1+x)^{2n}}\\ &=\frac12\sum_{n=2}^{\infty}nz^n-\frac12\sum_{n=2}^{\infty}z^n-\sum_{n=1}^{\infty}nz^{2n}+\frac12\sum_{n=1}^{\infty}z^{2n}\\ &=\frac12\left[-z+\sum_{n=1}^{\infty}nz^n\right]-\frac12\left[-1-z+\sum_{n=0}^{\infty}z^n\right]-\sum_{n=1}^{\infty}n(z^2)^{n}+\frac12\sum_{n=0}^{\infty}(z^2)^{n}-\frac12\\ &=\frac12\left[-z+\frac{z}{(1-z)^2}\right]-\frac12\left[-1-z+\frac{1}{1-z}\right]-\frac{z^2}{(1-z^2)^2}+\frac12\frac{1}{1-z^2}-\frac12\\ &=\frac{z^2}{2(1-z)^2}+\frac{1-3z^2}{2(1-z^2)^2}-\frac12\\ &=\frac{z^3}{(1-z^2)^2}\\ &=\frac{x+1}{x^2(x+2)^2}. \end{align}$$

The interval of convergence corresponds to $|z|<1$, or $\frac{1}{|1+x|}<1\iff (x>0)\lor(x<-2)$.