<s>[INST] Find the sum $\sum\limits_{k=0}^\infty (-2)^k\frac{k+2}{k+1}x^k$ $$\displaystyle\sum_{k=0}^\infty (-2)^k\dfrac{k+2}{k+1}x^k.$$
I showed that this series converges when $|x|<\dfrac{1}{2}$ because $$\displaystyle\lim_{k\to\infty}\left|\dfrac{a_{k+1}}{a_k}\right|=2|x|.$$
Now I have to find the sum result. I tried so far trying to combine $$\ln(x+1)=\displaystyle\sum_{k=0}^\infty \dfrac{(-1)^kx^{k+1}}{k+1}$$ and $$\arctan(x)=\displaystyle\sum_{k=0}^\infty \dfrac{(-1)^kx^{2k+1}}{2k+1},$$ and trying many substitutions with multiples of $x$, $x^2$ and got close to the result, but $\dfrac{k+2}{k+1}$ confuses me.
Any help would be appreciated.
 [/INST] So we want to find the value of
$$\sum_{k=0}^\infty \frac{(k+2)(-2x)^k}{k+1}$$
As you already determined, we know
$$\ln (x+1)=\sum_{k=0}^\infty \frac{(-1)^kx^{k+1}}{k+1}$$
If you multiply both sides by $x$ you get
$$x\ln (x+1)=\sum_{k=0}^\infty \frac{(-1)^k x^{k+2}}{k+1}$$
Now, the interesting step is to take the derivative of both sides.
$$\ln (x+1)+\frac{x}{x+1}=\sum_{k=0}^\infty \frac{k+2}{k+1}(-1)^kx^{k+1}$$
Now divide both sides by $x$ to get
$$\frac{1}{x}\ln (x+1)+\frac{1}{x+1}=\sum_{k=0}^\infty \frac{k+2}{k+1}(-x)^k$$
Substitute $2x$ into $x$ to get,
$$\ln (2x+1)+\frac{1}{2x+1}=\sum_{k=0}^\infty \frac{k+2}{k+1}(-2x)^k$$
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