The sum of the series 112-123+134- up to is equal to:

Question

The sum of the series $\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}-\cdots$ up to $\infty$ is equal to:

Accepted Answer

$T_n=\dfrac{(-1)^{n+1}}{n(n+1)}=(-1)^{n+1}\left(\dfrac1n-\dfrac{1}{n+1}\right)$. Writing out and regrouping:
$$S=\left(1-\frac12\right)-\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)-\cdots=1-2\left(\frac12-\frac13+\frac14-\cdots\right).$$

Using $\log(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots$ at $x=1$: $\log2=1-\dfrac12+\dfrac13-\cdots$, so $\dfrac12-\dfrac13+\dfrac14-\cdots=1-\log2$.

$$S=1-2(1-\log2)=2\log2-1=\log4-\log e=\log_e\frac4e.$$

Solutions · 1