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Find the value of $\displaystyle\sum_{n=1}^\infty \frac{(-1)^n}{n}$.

I have no idea how to start. I've tried listing out a few terms and I still can't find a pattern or anything. Any whatsoever help is appreciated!

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  • $\begingroup$ Hint: Consider the Taylor expansion of $\log (1+x)$. $\endgroup$ – lulu Feb 22 '17 at 22:23
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Hints: for $\;|x|<1\;$

$$\frac1{x(1+x)}=\sum_{n=0}^\infty(-1)^nx^{n-1}\implies \log\frac x {1+x}=\sum_{n=0}^\infty\frac{(-1)^n x^n}n\;\;\ldots$$

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I thought it might be instructive to present an approach that relies on only straightforward arithmetic and the identification of a Riemann sum. To that end, we now proceed.

Note that we can write

$$\begin{align}\sum_{n=1}^{2N} \frac{(-1)^{n-1}}{n}&=\sum_{n=1}^N \left( \frac{1}{2n-1}-\frac{1}{2n}\right)\\\\ &=\sum_{n=1}^N\left( \frac{1}{2n-1}+\frac{1}{2n}\right)-2\sum_{n=1}^N\frac{1}{2n}\\\\ &=\sum_{n=1}^{2N}\frac1n -\sum_{n=1}^N\frac1n\\\\ &=\sum_{n=N+1}^{2N}\frac1n\\\\ &=\sum_{n=1}^N\frac{1}{N+n}\\\\ &=\frac1N \sum_{n=1}^N\frac1{1+n/N}\tag 1 \end{align}$$

Recognizing that the limit of $(1)$ is the Riemann sum for $\int_0^1 \frac1{1+x}\,dx=\log(2)$, we find the limit of the alternating harmonic series is $\log(2)$.

Tools used: Arithmetic and Riemann Sums

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  • $\begingroup$ @ahmeds.attaalla Yes. Edited $\endgroup$ – Mark Viola Feb 23 '17 at 1:28
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Note that,

$$\int_{0}^{1} x^{n-1} dx=\frac{1}{n}$$

So that,

$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$$

$$=\sum_{n=1}^{\infty} \left( \int_{0}^{1} x^{n-1} dx \right)(-1)^n$$

Interchange the sum and integral.

$$=\int_{0}^{1} \sum_{n=1}^{\infty} x^{n-1}(-1)^{n-1} (-1) dx$$

$$=-\int_{0}^{1} \sum_{n=1}^{\infty} x^{n-1}(-1)^{n-1} dx$$

$$=-\int_{0}^{1} \sum_{n=0}^{\infty} (-x)^n dx$$

Recognize the geometric series to get,

$$=-\int_{0}^{1} \frac{1}{1+x} dx$$

$$=-\ln 2$$

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