Inspired by Patrick Corn

This solution is devoted to guys who don't know about $\text{Li}$ (including me)

$\displaystyle \sum _{ n=1 }^{ \infty }{ \dfrac { { H }_{ n }^{ (2) } }{ { 2 }^{ n } } } =2\times \sum _{ 1 }^{ \infty }{ \dfrac { 1 }{ { k }^{ 2 }{ 2 }^{ k } } } -\dfrac { 1 }{ { 2 }^{ n } } \sum _{ 1 }^{ \infty }{ \dfrac { 1 }{ { k }^{ 2 } } } =2\times \sum _{ 1 }^{ \infty }{ \dfrac { 1 }{ { k }^{ 2 }{ 2 }^{ k } } }$

Now, consider this sum

$\displaystyle \sum _{ 1 }^{ \infty }{ \dfrac { 1 }{ { k }^{ 2 }{ a }^{ k } } }=I(a)$

$\displaystyle \dfrac { d }{ da } I(a)=\dfrac { d }{ da } \sum _{ 1 }^{ \infty }{ \dfrac { 1 }{ { k }^{ 2 }a^{ k } } } =-\sum _{ 1 }^{ \infty }{ \dfrac{ 1 }{ { k }a^{ k+1 } } } =\dfrac { -1 }{ a } \sum _{ 1 }^{ \infty }{ \dfrac { 1 }{ { k }a^{ k } } }$

$\displaystyle \frac { d }{ da } I(a)=\dfrac { -1 }{ a } \sum _{ 1 }^{ \infty }{ \dfrac { 1 }{ { k }a^{ k } } } =\dfrac { 1 }{ a } \ln {( 1-\dfrac { 1 }{ a } ) }$

$\displaystyle \dfrac { d }{ da } I(a)=\dfrac { \ln {( a-1) } -\ln { a } }{ a }$

Now, let's integrate it!

$\displaystyle \int { \dfrac { d }{ da } I(a)da } =\int { \dfrac { \ln {( a-1) } -\ln { a } }{ a } da }$

But we know that $I(a)=\zeta(2)$ when a=1 and we want I(2) , hence,

$\displaystyle \int _{ a=1 }^{ 2 }{ \dfrac { d }{ da } I(a)da } =\int _{ a=1 }^{ 2 }{ \dfrac { \ln {( a-1) } -\ln { a } }{ a } da } =I(2)-\zeta (2)$

$\displaystyle \int _{ a=1 }^{ 2 }{ \dfrac { \ln { (a-1) } -\ln { a } }{ a } da } =\int _{ a=1 }^{ 2 }{ \dfrac { \ln { (a-1) } }{ a } da } -\int _{ a=1 }^{ 2 }{ \dfrac { \ln { a } }{ a } da } =\int _{ a=1 }^{ 2 }{ \dfrac { \ln {( a-1) } }{ a } da } -\dfrac { { (\ln { a } ) }^{ 2 } }{ 2 } =\int _{ a=1 }^{ 2 }{ \dfrac { \ln {( a-1 )} }{ a } da } -\dfrac { { (\ln { 2 } ) }^{ 2 } }{ 2 }$

Now, let's look at this integral.

$\displaystyle \int _{ a=1 }^{ 2 }{ \dfrac { \ln {( a-1) } }{ a } da } =\ln { (a-1) } \ln { a } -\int _{ a=1 }^{ 2 }{ \dfrac { \ln { a } }{ (a-1) } da } =-\int _{ a=1 }^{ 2 }{ \dfrac { \ln { a } }{ a-1 } da }$

From $\displaystyle \ln { a } =\sum _{ 1 }^{ \infty }{ \dfrac { { (a-1) }^{ n }{ (-1) }^{ n+1 } }{ n } }$

$\displaystyle -\int _{ a=1 }^{ 2 }{ \dfrac { \sum _{ 1 }^{ \infty }{ \dfrac { { (a-1) }^{ n }{ (-1) }^{ n+1 } }{ n } } }{ a-1 } da } =-\int _{ a=1 }^{ 2 }{ \sum _{ 1 }^{ \infty }{ \dfrac { { (a-1) }^{ n-1 }{ (-1) }^{ n+1 } }{ n } } da }$

$\displaystyle -\int _{ a=1 }^{ 2 }{ \sum _{ 1 }^{ \infty }{ \dfrac { { (a-1) }^{ n-1 }{ (-1) }^{ n+1 } }{ n } } da } =-\sum _{ 1 }^{ \infty }{ \dfrac { { (-1) }^{ n+1 } }{ { n }^{ 2 } } }$

$\displaystyle \sum _{ 1 }^{ \infty }{ \dfrac { { (-1) }^{ n+1 } }{ { n }^{ 2 } } } =1-\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ { 3 }^{ 2 } } -...$

$\displaystyle \zeta (2)-\sum _{ 1 }^{ \infty }{ \dfrac { { (-1) }^{ n+1 } }{ { n }^{ 2 } } } =2(\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ { 4 }^{ 2 } } +\dfrac { 1 }{ { 6 }^{ 2 } } ...)=\dfrac { 1 }{ 2 } \zeta (2)$

$\displaystyle \sum _{ 1 }^{ \infty }{ \dfrac { { (-1) }^{ n+1 } }{ { n }^{ 2 } } } =\dfrac { 1 }{ 2 } \zeta (2)$

Combine these results we get

$\displaystyle I(2)-\zeta(2)=-\dfrac{\zeta(2)}{2}-\frac{(\ln(2))^{2}}{2}$ or $I(2)=\dfrac{\zeta(2)}{2}-\dfrac{(\ln(2))^{2}}{2}$

Hence, the summation above reduces to

$\displaystyle \sum _{ 1 }^{ \infty }{ \dfrac { { H }_{ n }^{ (2) } }{ { 2 }^{ n } } } =2\left( \dfrac{\zeta(2)}{2}-\dfrac{(\ln(2))^{2}}{2} \right)=\dfrac{\pi^{2}}{6}-(\ln(2))^{2}$

Then,

$\displaystyle a+b+c+d=2+6+2+2=\boxed{12}$

Reversing the order of summation $\begin{array}{rcl} \displaystyle \sum_{n=1}^\infty \frac{H_n^{(2)}}{2^n} & = & \displaystyle \sum_{n=1}^\infty \sum_{k=1}^n \frac{1}{k^22^n} \; = \; \displaystyle \sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{k^22^n} \; = \; \displaystyle \sum_{k=1}^\infty \frac{1}{k^22^{k-1}} \\ & = & \displaystyle 2\mathrm{Li}_2(\tfrac12) \; =\; \displaystyle \tfrac16\pi^2 - (\ln 2)^2 \end{array}$ making the answer $2+6+2+2 = \boxed{12}$ .