How find the prime $p$ such $p\mid\sum_{k=1}^{p+2}\frac{T_{k}}{k+1}$ 
let $k\in N^{+}$, and such
  $$T_{k}=\sum_{i=1}^{k}\dfrac{1}{i\cdot 2^i}$$
  Find all prime number $p$, such that
  $$p\mid\sum_{k=1}^{p+2}\dfrac{T_{k}}{k+1}$$

I think this problem maybe use this  Abel transformation:see http://en.wikipedia.org/wiki/Summation_by_parts
let $$a_{n}=T_{n},b_{n}=\dfrac{1}{n+1},H_{n}=1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}$$
then
$$\sum_{k=1}^{p+2}\dfrac{T_{k}}{k+1}=T_{p+2}\cdot H_{p+2}-\sum_{k=1}^{p+1}H_{k}\dfrac{1}{k\cdot 2^k}$$
then I can't  Continue
 A: What I get from partial summation is:
$$\begin{eqnarray*}I_p=\sum_{k=1}^{p+2}\frac{1}{k+1}\sum_{i=1}^{k}\frac{1}{i\,2^i}&=&(H_{p+3}-1)T_{p+2}-\sum_{k=1}^{p+1}\frac{H_{k+1}-1}{(k+1)2^{k+1}}\\&=&H_{p+3}T_{p+2}-\sum_{k=1}^{p+2}\frac{H_k}{k\,2^k}.\tag{1}\end{eqnarray*}$$
Now, it is well-known that $H_{p-1}\equiv 0\pmod{p^2}$, and by this paper of Zhi-Wei Sun we know that:
$$\sum_{k=1}^{p-1}\frac{H_k}{k\, 2^k}\equiv 0\pmod{p},\tag{2}$$
so $p\,|\,I_p$ is equivalent to:
$$ p\; | \left(\frac{1}{p}+\frac{1}{p+1}+\frac{1}{p+2}+\frac{1}{p+3}\right)T_{p+2}-\frac{H_p}{p\,2^p}-\frac{H_{p+1}}{(p+1)\,2^{p+1}}-\frac{H_{p+2}}{(p+2)\,2^{p+2}},$$
or:
$$ p\; | \left(\frac{1}{p}+\frac{1}{p+1}+\frac{1}{p+2}+\frac{1}{p+3}\right)T_{p+2}-\frac{H_p}{2p}-\frac{H_{p+1}}{4}-\frac{H_{p+2}}{16},$$
$$ p\; | \left(\frac{1}{p}+\frac{1}{p+1}+\frac{1}{p+2}+\frac{1}{p+3}\right)T_{p+2}-\frac{1}{2p^2}-\frac{H_{p+1}}{4}-\frac{H_{p+2}}{16},$$
$$ p\; | \left(\frac{1}{p}+\frac{11}{6}\right)\left(T_{p-1}+\frac{1}{2p}+\frac{5}{16}\right)-\left(\frac{1}{2p^2}+\frac{5}{16p}+\frac{3}{32}\right),$$
$$ p\; | \frac{11}{6}\left(T_{p-1}+\frac{1}{2p}\right)+\frac{T_{p-1}}{p}+\frac{23}{48},\tag{3}$$
hence to solve the problem it suffices to study the residue classes of $T_{p-1}\pmod{p^2}$.
We must have:
$$ p^2 \mid  (48+88p) T_{p-1}+(23p+44), $$
from which
$$ 12 T_{p-1} + 11 \equiv 0 \pmod{p}$$
follows. The calculation of $T_{p-1}\pmod{p}$ occurs in the problem N7 of the IMO-2011 shortlist. 
It is shown that:
$$ T_{p-1}\equiv \frac{1-\frac{1}{2^{p-1}}}{p}\pmod{p},$$
so the only primes for which we may have divisibility are the ones for which:
$$ 2^{p-1}+\frac{11}{12}p 2^{p-1}\equiv 1\pmod{p^2}.$$
Anyway, we require a closed expression for $T_{p-1}\pmod{p^2}$ in order to really close the problem. 
I bet that a key element is the following identity:
$$\frac{1}{p}\binom{p}{k}\equiv (-1)^{k-1}\frac{1+pH_{k-1}}{k}\pmod{p^2},\tag{4}$$
from which we get:
$$ T_{p-1}\equiv\frac{1-\frac{1}{2^{p-1}}}{p}+p\cdot\sum_{k=1}^{p-1}\frac{1}{k^2 2^k}\pmod{p^2}.\tag{5}$$
