<s>[INST] 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?
 [/INST] If you use polygamma functions
$$S_1=\sum\limits_{k=1}^{n} \frac{1}{2k+1}=\frac{1}{2} \left(\psi ^{(0)}\left(n+\frac{3}{2}\right)-\psi
   ^{(0)}\left(\frac{3}{2}\right)\right)$$
$$S_2=\sum\limits_{k=1}^{n} \frac{1}{2k+2}=\frac{1}2\sum\limits_{k=1}^{n} \frac{1}{k+1}=\frac{1}{2} (\psi ^{(0)}(n+2)+\gamma -1)$$ making 
$$\sum\limits_{k=1}^{n} \left(\frac{1}{2k+1} - \frac{1}{2k+2}\right) = S_1-S_2$$ $$S_1-S_2=\frac{1}{2} \left(\psi ^{(0)}\left(n+\frac{3}{2}\right)-\psi
   ^{(0)}\left(\frac{3}{2}\right)\right)-\frac{1}{2} (\psi ^{(0)}(n+2)+\gamma -1)$$
If you look here $$H_n=\gamma +\psi ^{(0)}(n+1)$$ making
$$S_1-S_2=\frac{1}{2} \left(H_{n+\frac{1}{2}}-H_{n+1}-H_{\frac{1}{2}}+1\right)$$
and $1-H_{\frac{1}{2}}=\log (4)-1$ and then the result.
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