# Show that $\ln\frac23=\sum\limits_{n=1}^\infty\frac{(-1)^n}{2^nn}$

I am working on the following problem,

I have managed to first prove that the series is convergent, using the conditions for the alternating series test, although am unsure how to find the exact sum. I have attempted a method similar to telescoping series although this did not work. A previous part of the exercise was to find the following.

I am wondering if it is possible to use this result to prove the sum of the first series is ln(2/3), or if there is a method of calculating the sum i am unaware of. I am aware of the alternating series approximation however this questions asks to prove the exact value so i do not think it is applicable. Am i missing something major?

Thanks

• Your series is $\ln 2 + f(-1) = \ln 2 + f(0) - \int_{-1}^0 f'(x)\,dx$. – Daniel Fischer Sep 28 '16 at 14:53

You don't have to do complicated things:

$$\ln(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n-1}n x^n}{n}\qquad\text{for}-1<x<1.$$ In particular $$\ln\frac32=\sum_{n=1}^\infty\frac{(-1)^{n-1} }{2^n n}.$$

• A very minor slip: the exponent on the $(-1)$ should be $n-1$, so that the series starts $x-x^2/2+\cdots$. – Lubin Sep 28 '16 at 15:05
• Oh yes! Fixed. Thanks for pointing it! – Bernard Sep 28 '16 at 15:19
• And you may want to change $\ln(2/3)=-\ln(3/2)=-\ln(1+1/2)$ – S. Y Sep 28 '16 at 21:01
• @S. Y.: I suppose anyone coming on this site is able to do it him/herself. – Bernard Sep 28 '16 at 21:15

Massive Hint: Since $\frac{1}{n 2^n}=\int_{0}^{1/2}x^{n-1}\,dx$, $$\sum_{n=1}^{N}\frac{(-1)^n}{n 2^n} = \int_{0}^{1/2}\sum_{n=1}^{N}(-1)^n x^{n-1}\,dx = \color{green}{-\int_{0}^{1/2}\frac{dx}{1+x}}+(-1)^N\color{blue}{\int_{0}^{1/2}\frac{x^N}{1+x}\,dx}$$ where the green integral equals $\color{green}{\log\frac23}$ and the blue integral is positive but bounded by $$\int_{0}^{1/2}x^N\,dx = \frac{1}{(N+1)2^{N+1}}.$$