Solving Summation $(2 + 1 \sum_{n=1}^{m} \frac{2}{n}) - (2 + \sum_{n=2}^{m-1} \frac{4}{n+1} + \frac{4}{m+1})$... I saw this in a textbook and I am not sure how to get from the LHS to the RHS:
$$\left(2 + 1 + \sum_{n=1}^{m} \frac{2}{n}\right) - \left(2 + \sum_{n=2}^{m-1} \frac{4}{n+1} + \frac{4}{m+1}\right) + \left(\sum_{n=1}^{m-2} \frac{2}{n+2} + \frac{2}{m+1} + \frac{2}{m+2}\right) = 1- \frac{2}{m+1} + \frac{2}{m+2}$$
If somebody could show me the intermediate steps, that would be really helpful.
 A: The series as written:
$$\left(2 + 1 + \sum_{n=1}^{m} \frac{2}{n}\right) - \left(2 + \sum_{n=2}^{m-1} \frac{4}{n+1} + \frac{4}{m+1}\right) + \left(\sum_{n=1}^{m-2} \frac{2}{n+2} + \frac{2}{m+1} + \frac{2}{m+2}\right) $$
gives
$$ S_{m} = 1 - \frac{2}{m+1} + \frac{2}{m+2} + 2 \, f_{m} $$
where
\begin{align}
f_{m} &= \sum_{n=1}^{m} \frac{1}{n} - \sum_{n=2}^{m-1} \frac{2}{n+1} + \sum_{n=1}^{m-2} \frac{1}{n+2} \\
&= \sum_{n=1}^{m} \frac{1}{n} - \sum_{n=3}^{m} \frac{2}{n} + \sum_{n=3}^{m} \frac{1}{n} = \sum_{n=1}^{m} \frac{1}{n} - \sum_{n=3}^{m} \frac{1}{n} \\
&= 1 + \frac{1}{2}
\end{align}
and leads to
$$ S_{m} = 4 - \frac{2}{m+1} + \frac{2}{m+2}. $$
This is not the result presented.
In order to obtain the claimed result it would seem to be more like:
$$ S_{m} = 2 \, \left(1 + \sum_{n=1}^{m} \frac{1}{n} \right) - 2 \, \left(1 + \frac{2}{m+1} + \sum_{n=1}^{m-1} \frac{2}{n+1} \right) + 2 \, \left(\frac{1}{m+1} + \frac{1}{m+2} + \sum_{n=1}^{m-2} \frac{1}{n+2} \right) $$
which gives
\begin{align}
S_{m} &= 2 \, \left[ - \frac{1}{m+1} + \frac{1}{m+2} + \sum_{n=1}^{m} \frac{1}{n} - \sum_{n=2}^{m} \frac{2}{n} + \sum_{n=3}^{m} \frac{1}{n} \right] \\
&= 2 \, \left[ - \frac{1}{m+1} + \frac{1}{m+2} + \frac{1}{1} + \frac{1}{2} - \frac{2}{2} \right] \\
&= 1 - \frac{2}{m+1} + \frac{2}{m+2}.
\end{align}
