Sum $\sum_{n=1}^\infty (\log 2)^n \sum_{r=1}^n \frac {r^2}{r!(n-r)!},$ The general term of a sequence is given as
 $$\displaystyle T_n = (\log 2)^n \sum_{r=1}^n \frac {r^2}{r!(n-r)!},$$ $n\in \mathbb N$. 
How can I find an expression for the below given summation?
$$ \sum_{k=1}^\infty T_k$$
Multiplying and dividing by $n!$ the first expression is reduced to $$ T_n= \frac{(\log2)^n}{n!} \sum_{r=1}^{n} {^nC_r} r^2$$ 
$$\implies T_n=2^{n-2} (\log2)^n \frac{ (n+1)}{(n-1)!}$$
But I cannot think of any method to evaluate $\sum_{k=1}^{\infty} T_k$.
Is there some hidden telescoping I am missing?
Is it possible that this sum to infinity be reduced to an integral? If so, how can it be done?

We have
$$e^x = \sum_{n=0}^{\infty} \frac{1}{n!} x^n$$
This implies that
$$e^x = \frac{d}{dx}e^x = \sum_{n=1}^{\infty} \frac{1}{(n-1)!} x^{n-1}$$
Thus,
$$x^2 e^x = x^2 \frac{d}{dx}e^x = \sum_{n=1}^{\infty} \frac{1}{(n-1)!} x^{n+1}$$
Taking a second derivative gives
$$(2x+x^2) e^x = \frac{d}{dx}(x^2 e^x) = \sum_{n=1}^{\infty} \frac{n+1}{(n-1)!} x^{n}$$
Now,
$$\begin{align}
\sum_{n=1}^{\infty} T_n & = \sum_{n=1}^{\infty} 2^{n-2} (\ln 2)^n \frac{n+1}{(n-1)!} \\
& = \frac14 \sum_{n=1}^{\infty} \frac{n+1}{(n-1)!}  (2 \ln 2)^n\\
\end{align}$$
Thus,
$$\sum_{n=1}^{\infty} T_n = \frac14 \left.(2x+x^2) e^x \right|_{x=2\ln 2} = \frac14(2\cdot 2\ln 2+(2\ln 2)^2)e^{2\ln 2} = (\ln 2 + (\ln 2)^2) \cdot 4$$
