Infinite series and logarithm

Is it true that:

$$\log_e 2 = \frac12 + \frac {1}{1\cdot2\cdot3} + \frac {1}{3\cdot4\cdot5}+ \frac{1}{5\cdot6\cdot7}+ \ldots$$

It was one of my homeworks . Thanks!

Answer

This calculation may look a bit roundabout, but it accurately reflects my thought processes (otherwise known as directed tinkering!) in solving the problem.

Start with the Maclaurin series $$\ln(1+x)=\sum_{n\ge 1}(-1)^{n+1}\frac{x^n}n\;,$$ which is valid for $-1

$$\begin{align*} \ln 2&=\sum_{n\ge 1}\frac{(-1)^{n+1}}n\\ &=1-\frac12+\frac13-\frac14\pm\ldots\\ &=\left(1-\frac12\right)+\left(\frac13-\frac14\right)+\left(\frac15-\frac16\right)+\ldots\\ &=\sum_{n\ge 1}\left(\frac1{2n-1}-\frac1{2n}\right)\\ &=\sum_{n\ge 1}\frac1{2n(2n-1)}\;. \end{align*}$$

The series in the problem is

$$\begin{align*} \frac12+\frac1{1\cdot2\cdot3}+\frac1{3\cdot4\cdot5}+\frac1{5\cdot6\cdot7}+\ldots&=\frac12+\sum_{n\ge 1}\frac1{(2n-1)(2n)(2n+1)}\\ &=\frac12+\sum_{n\ge 1}\left(\frac1{2n(2n-1)}\cdot\frac1{2n+1}\right)\\ &=\frac12+\sum_{n\ge 1}\frac1{2n(2n-1)}\left(1-\frac{2n}{2n+1}\right)\\ &=\frac12+\sum_{n\ge 1}\frac1{2n(2n-1)}-\sum_{n\ge 1}\left(\frac1{2n(2n-1)}\cdot\frac{2n}{2n+1}\right)\\ &=\frac12+\sum_{n\ge 1}\frac1{2n(2n-1)}-\sum_{n\ge 1}\frac1{(2n-1)(2n+1)}\\ &=\frac12+\sum_{n\ge 1}\frac1{2n(2n-1)}-\frac12\sum_{n\ge 1}\left(\frac1{2n-1}-\frac1{2n+1}\right)\;. \end{align*}$$

Now notice that $$\sum_{n\ge 1}\left(\frac1{2n-1}-\frac1{2n+1}\right)$$ telescopes, so it can be evaluated easily; do that, and you’ll have the desired result. (You also have to justify the various manipulations that rearrange the terms of the series in the last long calculation, but that’s not a problem: everything in that calculation is absolutely convergent.)