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!

$$a_n = \dfrac1{(2n-1)2n(2n+1)} = \dfrac12 \left( \dfrac1{2n-1} - \dfrac1{2n} - \dfrac1{2n} + \dfrac1{2n+1}\right)$$
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}
