A perfectly natural question ....

$S=\displaystyle\sum_{n=1}^{\infty} \frac{1}{2^{n}*(n-\frac{1}{2})} = \sqrt{2}\displaystyle\sum_{n=1}^{\infty} \frac {(\dfrac {1}{\sqrt{2}})^{(2n-1)}}{(2n-1)}$

Now,we know that,

$\log {(1+x)} = \displaystyle\sum_{n=1}^{\infty}(-1)^{(n-1)} \frac {x^n}{n}$

Also,

$-\log{(1-x)} =\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n}$

Adding both gives us,

$\log{(\frac {1+x}{1-x})}*{\frac{1}{2}}=\displaystyle\sum_{n=1}^{\infty}\frac{x^{(2n-1)}}{2n-1}$

Therefore,

$S=\sqrt{2}\displaystyle\sum_{n=1}^{\infty} \frac {(\dfrac {1}{\sqrt{2}})^{(2n-1)}}{(2n-1)} = \frac {1}{2} \sqrt {2} {\log (\frac {1+\frac {1}{\sqrt{2}}}{1-\frac {1}{\sqrt{2}}})}$

Rationalizing,the radical,we get,

$S= \frac {1}{2} \sqrt {2} {\log {(1+\sqrt{2})}^2}=\sqrt {2} {\log {(1+\sqrt{2})}}$

The sum is $\sqrt{2}*\ln(1 + \sqrt{2})$ , making $a^{b+c} = 2^{1+2} = \boxed{8}$ .

I debated whether to post this as a question or a note, since I'm still working on a proof. It's such a beautiful result, though, that I wanted to get it posted regardless and see if the community has any ideas on an elegant proof.

i used integrals on (n-1/2) via 1/A = integral 0 to inf e^(-Ax)dx , A>0