Sums and harmonic series I found the solution of series on Wolfram Alpha 
http://www.wolframalpha.com/input/?i=sum+1%2F(2k%2B1)%2F(2k%2B2)+from+1+to+n
$ \sum\limits_{k=1}^{n} \left(\frac{1}{2k+1} - \frac{1}{2k+2}\right) = \sum\limits_{k=1}^{n} \frac{1}{(2k+1)(2k+2)} = \frac{1}{2} \left(H_{n+\frac{1}{2}} - H_{n+1} -1 + \text{ln}(4)\right)$
Can someone tell how to prove this in the form of Harmonic numbers?

This is not an answer to the specific question but a long comment which gives a derivation of a much simpler formula for the sum in question. 
Let
$$s = \sum _{k=1}^n \left(\frac{1}{2 k+1}-\frac{1}{2 k+2}\right)$$
Adding and subtracting a sum of even terms we get for $s$
$$\begin{array}
&=& \sum _{k=1}^n \left(\frac{1}{2 k}+\frac{1}{2 k+1}\right)-\sum _{k=1}^n \left(\frac{1}{2 k}+\frac{1}{2 k+2}\right)\\
=& (\frac{1}{2}+\frac{1}{3}+ ... + \frac{1}{2n+1})-\frac{1}{2}(1+\frac{1}{2}+...+\frac{1}{n}) -\frac{1}{2}(\frac{1}{2}+\frac{1}{3}+...+ \frac{1}{n+1})\\
=&(H_{2n+1}-1) -\frac{1}{2}H_n -\frac{1}{2}(H_{n+1}-1)\\
=&H_{2n+1} -\frac{1}{2}(H_n +H_{n+1})-\frac{1}{2}
\end{array}
$$
Those who wish can simplifiy this further using the relation $H_{n+1}=H_n + \frac{1}{n+1}$.
