Calculate the following sum

Question

Calculate the following sum $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)}$$ I Tried decomposing the denominator but still seem to be stuck thanks to the numerator

Answer 1

Your series is absolutely convergent by the p-test, and by integrating twice $$ \sum_{n\geq 1}(-1)^{n+1}x^{n-1} = \frac{1}{1+x} $$ we get $$ \sum_{n\geq 1}\frac{(-1)^{n+1} x^{n+1}}{n(n+1)} = (1+x)\log(1+x)-x $$ for any $x\in(-1,1)$. Since the RHS is continuous in a neighbourhood of $x=1$ and the original series is absolutely convergent, such identity holds at $x=1$ too, leading to: $$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n(n+1)} = \color{red}{\log\left(\frac{4}{e}\right)}.$$

Answer 2

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)}=-1+2\sum_{r=1}^\infty\dfrac{(-1)^r}r=-1+2\ln2$$

See "Proof" that $\log2=0$ using the expansion of $\log(1+x)$

Answer 3

For the proof of $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$=$\ln2$ , see here .So expanding the terms with Crostul's hint, to see that the sum equals $2\ln2-1$

Answer 4

$$\sum_{n=1}^\infty\frac{-1^{n+1}}{n(n+1)} = \frac12 - \frac16 + \frac{1}{12} - \frac{1}{20} + \frac{1}{30} - \frac{1}{42} + ...$$ = $$\frac12(1 + \frac16 + \frac{1}{15} + \frac{1}{28} + ...) - \frac12(\frac13 + \frac{1}{10} + \frac{1}{21} + \frac{1}{36} + ...)$$ The sum of the reciprocals of triangle numbers converges to $2$, but the sum of the $1st, 3rd, 5th...$ reciprocals $\approx$ $1.38629$, while the sum of the $2nd, 4th, 6th... \approx .613706$. Thus the difference between the half-sums $\approx .386292$.