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

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 $x \in\{0, -2\}$.
