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Find the sum of the series $\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2^kk}$

Whether I can find some $x$ such that $f(x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2^kk}$ ,then find the closed form of $f(x)$.

And substitute x into the closed of $f(x)$ to find out the sum?

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3 Answers 3

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You can manipulate a geometric series to look like that series.

$$\sum_n^\infty q^n=\frac{1}{1-q}$$

Substituting $-x=q$

$$\sum_n^\infty (-1)^n x^{n}=\frac{1}{1+x}$$

Dividing both sides by $x$.

$$\sum_n^\infty (-1)^n x^{n-1}=\frac{1}{x(1+x)}$$

You can now integrate and substitute $x=1/2$, can you finish this?

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  • $\begingroup$ I substitute $x=1/2$ → $\sum_{n=1}^\infty \frac{(-1)^n}{2^{n-1}}$.How to integrate it to $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n2^n}$? $\endgroup$
    – Wang Chen
    Apr 13, 2017 at 16:19
  • $\begingroup$ You integrate it first to look like $\sum_{n=1}^\infty (-1)^n \frac{x^n}{n}$ then you substitute, you still have the wrong index for the $-1$ but you should know how to fix that. $\endgroup$
    – Rab
    Apr 13, 2017 at 16:22
  • $\begingroup$ OK, I think I get it. Thanks $\endgroup$
    – Wang Chen
    Apr 13, 2017 at 16:36
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For any $x$ such that $|x|<1$ we have $\sum_{n\geq 1}\left(-\frac{x}{2}\right)^{n-1}=\frac{1}{1+\frac{x}{2}}$, and by applying $\frac{1}{2}\int_{0}^{1}\left(\ldots\right)\,dx$ to both sides we get: $$ \sum_{n\geq 1}\frac{(-1)^{n-1}}{n 2^{n}} = \int_{0}^{1}\frac{dx}{2+x} = \log\left(\frac{3}{2}\right).$$

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$$\sum_{k=1}^\infty\dfrac{(-1)^{k-1}}{k2^k}=-\sum_{k=1}^\infty\dfrac{(-1/2)^k}k$$

Now for $\displaystyle-1\le x<1,\ln(1-x)=-\sum_{k=1}^\infty\dfrac{x^k}k$

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