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!
 A: $$a_n = \dfrac1{(2n-1)2n(2n+1)} = \dfrac12 \left( \dfrac1{2n-1} - \dfrac1{2n} - \dfrac1{2n} + \dfrac1{2n+1}\right)$$
and $$a_0 = \dfrac12$$
Hence,
\begin{align}
\sum_{k=0}^n a_k & = \dfrac12 + \dfrac12 \left(\dfrac11 - \dfrac12 - \dfrac12 + \dfrac13 +  \dfrac13 - \dfrac14 - \dfrac14 + \dfrac15 + \cdots + \dfrac1{2n-1} - \dfrac1{2n} - \dfrac1{2n} + \dfrac1{2n+1}\right)\\
& = \dfrac12 \left(1 + 1 - \dfrac12 - \dfrac12 + \dfrac13 + \dfrac13+ \cdots + \dfrac1{2n-1} + \dfrac1{2n-1} - \dfrac1{2n} - \dfrac1{2n} + \dfrac1{2n+1} \right)\\
& = 1 - \dfrac12 + \dfrac13 - \dfrac14 + \dfrac15 \mp \cdots + \dfrac12\dfrac{1}{2n+1}
\end{align}
A: 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 $-1<x\le 1$, and set $x=1$:
$$\begin{align*}
\ln 2&=\sum_{n\ge 1}\frac{(-1)^{n+1}}n\\
&=1-\frac12+\frac13-\frac14\pm\ldots\\
&=\left(1-\frac12\right)+\left(\frac13-\frac14\right)+\left(\frac15-\frac16\right)+\ldots\\
&=\sum_{n\ge 1}\left(\frac1{2n-1}-\frac1{2n}\right)\\
&=\sum_{n\ge 1}\frac1{2n(2n-1)}\;.
\end{align*}$$
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.)
