'e'eesh! What a series!

The integral turns out to be $\displaystyle \dfrac{ 1 - 2^{-(k+1)} }{ (k + 1) \ln 2 }$ .

So, the inner sum is: $\displaystyle \dfrac{1}{\ln2} \sum_0^n k - 2^{-(k+1)} = \dfrac{1}{\ln2} \left( \dfrac{n(n+1)}2 - 1 + (\dfrac 1 2 )^{n+1} \right)$ .

Thus, the outer (and the most interesting) sum is: $\displaystyle \dfrac{1}{\ln2} \left( \sum_0^{\infty} \dfrac{n(n+1)}{2n!} - \dfrac 1 {n!} + \left(\dfrac 1 2 \right)\cdot \dfrac{ (0.5)^n } {n!} \right)$ .

The second and third terms sum up nicely to $-e$ and $0.5 \sqrt{ e }$ .

We're now left with: $\displaystyle \dfrac{1}{\ln2} \left( \left(\sum_1^\infty \dfrac{n+1}{2(n-1)!}\right) - e + \dfrac 12 \sqrt e \right)$ .

(note the change in the limits).

$\displaystyle \dfrac{1}{\ln2} \left( \left(\sum_1^\infty \dfrac{n+1}{2(n-1)!}\right) - e + \dfrac 12 \sqrt e \right)$

= $\displaystyle \dfrac{1}{\ln2} \left( \left(\sum_1^\infty \dfrac{n}{2(n-1)!} + \dfrac{1}{2(n-1)!} \right) - e + \dfrac 12 \sqrt e \right)$

= $\displaystyle \dfrac{1}{\ln2} \left( \left(\sum_1^\infty \dfrac{n}{2(n-1)!}\right) - \dfrac 12 e + \dfrac 12 \sqrt e \right)$ .

Now, write the $n$ in the numerator as $n-1+1$ , and then the sum $\displaystyle \sum_1^{\infty} \dfrac{n}{2(n-1)!}$ easily evaluates to $e$ .

Putting all things together, the answer is: $\dfrac{ e + \sqrt e } {\ln 4 }$ .