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As stated in the title, I am trying to prove that $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \log2$, I think I need to use the Geometric Series like this:

$$1+u+u^2+...+u^n = \frac{1-u^{n+1}}{1-u}$$

Then setting $u = -t$, rewrite it like so:

$$1-t+t^2-...+(-1)^n t^n + \frac{(-1)^{n+1}t^{n+1}}{1+t}= \frac{1}{1+t}$$

But now I am not sure where to go next in order to get to my conclusion, any guidance and explanation would be appreciated, thanks!

  • 1
    Almost there. Integrate both sides and let $t=1$. In the process, take a moment to think about $+C$. – 2017-01-06
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    It is in the integration that I become confused I think, I have got this: $(1-\frac{1}{2}+...+\frac{(-1)^n}{n+1})$ for the first n terms but then I am not sure how best to integrate the final term, the RHS of course goes to $\log 2$ which is promising though? – 2017-01-06
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    Sorry, I wasn't looking closely enough. Integrate the infinite series (assuming absolute convergence) $1+u+u^2 + \cdots = 1/(1-u)$ and plug in $u=-1$. You're at the endpoint of the interval of convergence, so you have to so some thinking about convergence. Kobe's answer does this. – 2017-01-06
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    See also [Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $](http://math.stackexchange.com/q/716) and other questions [linked there](http://math.stackexchange.com/questions/linked/716). Found [using Approach0](https://approach0.xyz/search/?q=%24%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20%5Cfrac%7B(-1)%5E%7Bn%2B1%7D%7D%7Bn%7D%20%3D%20%5Clog%202%24&p=1). – 2017-01-06

3 Answers 3

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Take the integral from $0$ to $1$ on both sides of the equation, yielding

$$1 - \frac{1}{2} + \frac{1}{3} -\cdots + (-1)^n\frac{1}{n+1} + (-1)^{n+1}\int_0^1 \frac{t^{n+1}}{1+t}\, dt = \log 2$$

We want to show

$$\lim_{n\to\infty} (-1)^{n+1}\int_0^1 \frac{t^{n+1}}{1+t}\, dt = 0\tag{*}$$

To do so, we use the inequality $1 + t \ge 1$ for $t \ge 0$ to obtain

$$\left\lvert (-1)^{n-1} \int_0^1 \frac{t^{n+1}}{1+t}\, dt\right\rvert = \int_0^1 \frac{t^{n+1}}{1+t}\, dt \le \int_0^1 t^{n+1}\, dt = \frac{1}{n+2}$$

Since $\lim_{n\to \infty} \frac{1}{n+2} = 0$, (*) holds.

2

hint

Let $f(x)=\ln(1+x)$. $f$ is $C^{\infty}$ near $0$. thus for $n>0, \exists c\in(0,1)\;:\;$

$$f(1)=f(0)+\sum_{k=1}^n\frac{1}{k!}f^{(k)}(0)+\frac{1}{(n+1)!}f^{(n+1)}(c)$$

but $f^{(n+1)}(c)=(-1)^{n}\frac{n!}{(1+c)^{n+1}}$

and $0<\frac{1}{1+c}<1$ , so

$$\lim_{n\to+\infty}\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}=f(1)=\ln(2).$$

0

Fom the Taylor expansion of the natural logarithm

$$\ln(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}nx^n$$

With $x=1$, $\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=\ln(2)$