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%2F2
for $\alpha = 1$
$\sum\limits_{k=1}^{n/2} \left(\frac{1}{\left(2k-1+2^{1/\alpha}\right)^\alpha} - \frac{1}{\left(2k+2^{1/\alpha}\right)^\alpha}\right)=\frac{1}{2} \left(-H_{\frac{n}{2}+1} + H_\frac{n+1}{2} -1 + \text{ln}(4)\right)$
Can someone tell how to prove this in the form of Harmonic numbers?
A simpler expression, can be obtained via direct calculation. This avoids introducing irrational terms (such as $\log(4)$ in the solution from Wolfram Alpha).
Assuming $n$ is even (else the summation you provide is not defined), for $\alpha = 1$ the summation is equivalent to
\begin{align*} \sum_{k=1}^{n/2} \left( \frac{1}{2k+1} - \frac{1}{2(k+1)} \right) & = \sum_{k=1}^{n/2} \frac{1}{2k+1} - \frac12 \sum_{k=1}^{n/2} \frac{1}{k+1}\\ & = \sum_{k=1}^{n/2} \frac{1}{2k+1} - \frac12 \big(H_{n/2 +1} - 1 \big)\\ & = \sum_{k=1}^{n/2} \left( \frac{1}{2k+1} + \frac{1}{2k} - \frac{1}{2k} \right) - \frac12 \big(H_{n/2 +1} - 1 \big) \\ & = \sum_{k=1}^{n/2} \left( \frac{1}{2k+1} + \frac{1}{2k}\right) - \sum_{k=1}^{n/2} \frac{1}{2k}- \frac12 \big(H_{n/2 +1} - 1 \big) \\ & = \sum_{k=1}^{n/2} \left( \frac{1}{2k+1} + \frac{1}{2k}\right) - \frac12\sum_{k=1}^{n/2} \frac{1}{k}- \frac12 \big(H_{n/2 +1} - 1 \big) \\ & = \sum_{k=1}^{n/2} \left( \frac{1}{2k+1} + \frac{1}{2k}\right) - \frac12 H_{n/2} - \frac12 \big(H_{n/2 +1} - 1 \big) \\ & = \sum_{k=2}^{n+1} \frac1k - \frac12 H_{n/2}- \frac12 \big(H_{n/2 +1} - 1 \big) \\ & = (H_{n+1} -1) - \frac12 H_{n/2} - \frac12 \big(H_{n/2 +1} - 1 \big) \\ & =H_{n+1} - \frac12 \left(H_{n/2} + H_{n/2 + 1}-1 \right) \end{align*}
To test the answer, consider the arbitrary case $n = 16$
Left hand side. http://www.wolframalpha.com/input/?i=sum+1%2F(+(2k%2B1)++(2k%2B2)+)+from+1+to+8
Right hand side. http://www.wolframalpha.com/input/?i=H(n%2B1)+-1+-+1%2F2+H(n%2F2)+-+1%2F2+(H(n%2F2+%2B1)+-1)+with+n+%3D+16&rawformassumption=%7B%22FunClash%22,+%22H%22%7D+-%3E+%7B%22HarmonicNumber%22%7D