need to show sum of series equals 2log(2) - 1

[Euroman24]

I need show that

\(\displaystyle \sum \limit_{n=1} ^{\infty} \frac{(-1)^{n+1}}{n(n+1)} = 2 log(2) - 1\)

I know that I need to use partial sums.

Such that

\(\displaystyle \frac{(-1)^{n+1}}{n} + \frac{1}{(n+1)}\)

But where do I go from there?

Sincerely
Fred

 

[soroban]

Hello, Fred!

I have a solution, but it requires Recognition . . .

Show that: \(\displaystyle \L\:S\;=\;\sum \limit_{n=1} ^{\infty} \frac{(-1)^{n+1}}{n(n\,+\,1)} \;= \;2\cdot\ln 2 - 1\)

Since \(\displaystyle \L\,\frac{1}{n(n\,+\,1)}\:=\:\frac{1}{n}\,-\,\frac{1}{n\,+\,1}\)

we have:
\(\displaystyle \L\,S\;=\;\left(\frac{1}{1}\,-\,\frac{1}{2}\right)\,-\,\left(\frac{1}{2}\,-\,\frac{1}{3}\right)\,+\,\left(\frac{1}{3}\,-\,\frac{1}{4}\right)\,-\,\left(\frac{1}{4}\,-\,\frac{1}{5}\right)\,-\,\cdots\)

\(\displaystyle \L\;\;\;= \;1\,-\,\frac{1}{2}\,-\,\frac{1}{2}\,+\,\frac{1}{3}\,+\,\frac{1}{3}\,-\,\frac{1}{4}\,-\,\frac{1}{4}\,+\,\frac{1}{5}\,+\,\frac{1}{5}\,-\,\cdots\)

\(\displaystyle \L\;\;\;= \;1\,-\,\frac{2}{2}\,+\,\frac{2}{3}\,-\,\frac{2}{4}\,+\,\frac{2}{5}\,-\,\cdots\)


Add 1 and subtract 1:
\(\displaystyle \L S\;=\;\left(2\,-\,\frac{2}{2}\,+\,\frac{2}{3}\,-\,\frac{2}{4}\,+\,\frac{2}{5}\,-\,\cdots\right)\,-\,1\)

\(\displaystyle \L\;\;\;= \;2\underbrace{\left(1\,-\,\frac{1}{2}\,+\,\frac{1}{3}\,-\,\frac{1}{4}\,+\,\frac{1}{5}\,-\,\cdots\right)}\,-\,1\)
. . . . . . . . . . . . . . \(\displaystyle \text{This is }\ln 2\)


Therefore: \(\displaystyle \L\:S\;=\;2\cdot\ln 2\,-\,1\)