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?
On
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.
On
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 $H_{n+1}=H_n + \frac{1}{n+1}$.
We will make use of the following integral representation for the harmonic numbers $$H_x = \int^1_0 \frac{1 - t^x}{1 - t} \, dt. \tag1$$
Let $$S = \sum^n_{k = 1} \frac{1}{(2k + 1)(2k + 2)} = \sum^n_{k = 1} \left (\frac{1}{2k + 1} - \frac{1}{2k + 2} \right ).$$ Noting that $$\int^1_0 x^{2k} \, dx = \frac{1}{2k + 1} \quad \text{and} \quad \int^1_0 x^{2k + 1} \, dx = \frac{1}{2k + 2},$$ our sum can be rewritten as $$S = \sum^n_{k = 1} \int^1_0 (x^{2k} - x^{2k + 1}) \, dx = \int^1_0 (1 - x) \sum^n_{k = 1} x^{2k} \, dx.$$ As the finite sum appearing here is geometric, it can be summed. As $$\sum^n_{k = 1} x^{2k} = \frac{x^2 (1 - x^{2n})}{1 - x^2},$$ one has $$S = \int^1_0 \frac{x^2 ( 1 - x^{2n})}{1 + x} \, dx = \int^1_0 \left [\frac{x^2}{1 + x} - \frac{x^{2n + 2}}{1 + x} \right ] \, dx = I_1 - I_2.$$
The first integral is trivial. Here $$I_1 = \int^1_0 \frac{x^2}{1 + x} \, dx = \int^1_0 \left (x - 1 + \frac{1}{1 + x} \right ) \, dx = \left [\frac{x^2}{2} - x + \ln (1 + x) \right ]^1_0 = -\frac{1}{2} + \ln (2).$$
For the second integral \begin{align*} I_2 &= \int^1_0 \frac{x^{2n + 2}}{1 + x} \cdot \frac{1 - x}{1 - x} \, dx = \int^1_0 \frac{(1 - x) x^{2n + 2}}{1 - x^2} \, dx. \end{align*} Letting $x \mapsto \sqrt{x}$ gives \begin{align*} S &= \frac{1}{2} \int^1_0 \frac{x^{n + 1/2} - x^{n + 1}}{1 - x} \, dx\\ &= \frac{1}{2} \int^1_0 \frac{(1 - x^{n + 1}) - (1 - x^{n + 1/2})}{1 - x} \, dx\\ &= \frac{1}{2} \int^1_0 \frac{1 - x^{n + 1}}{1 - x} - \frac{1}{2} \int^1_0 \frac{1 - x^{n + 1/2}}{1 - x} \, dx\\ &= \frac{1}{2} \left (H_{n + 1} - H_{n + 1/2} \right ), \end{align*} where we in the last line we have made use of (1).
Thus $$\sum^n_{k = 1} \frac{1}{(2k + 1)(2k + 2)} = \ln (2) - \frac{1}{2} - \frac{1}{2} \left (H_{n + 1/2} - H_{n + 1} \right ),$$ as required.