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$$\displaystyle\sum_{k=0}^\infty (-2)^k\dfrac{k+2}{k+1}x^k.$$

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.

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    $\begingroup$ I would think of $(k+2)/(k+1)$ as $1 + 1/(k+1)$. Split the sum. Then you have $\sum (-2x)^k$ on the one hand (which you can handle) and $\sum (-2x)^k/(k+1)$, which might as well be $\sum (-y)^k/(k+1)$, which is $1/y$ times a natural log series. $\endgroup$
    – davidlowryduda
    Commented Dec 8, 2021 at 22:17
  • $\begingroup$ Thanks! I tried hard stuff instead of dividing that trivial sum haha. $\endgroup$
    – Fabrizio G
    Commented Dec 8, 2021 at 22:19

2 Answers 2

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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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    $\begingroup$ What a nice approach! Thank you very much! :-) I'll select this as best answer, closed. $\endgroup$
    – Fabrizio G
    Commented Dec 8, 2021 at 22:22
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Hint: observe that $\frac{k+2}{k+1} = 1+\frac{1}{k+1}$.

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    $\begingroup$ Thanks! Didn't notice this triviality. $\endgroup$
    – Fabrizio G
    Commented Dec 8, 2021 at 22:19

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