An olympiad number theory problem asking for prime numbers with a certain property Let $\displaystyle T_k=\sum^{k}_{j=1} \frac{1}{j2^j}$, where $k$ is a positive integer. Find all prime numbers $p$ such that $$\displaystyle\sum_{k=1}^{p-2}\frac{T_k}{k+1}\equiv 0 \pmod{p},$$ where $\frac{a}{b} \pmod{p}$ means $ab^{-1} \pmod{p}$ and $b^{-1}$ is the multiplicative inverse of $b \pmod{p}$.
This is an olympiad  problem asked by my previous teacher in my high school. It is so hard to me that the only thing I can do is to check whether the desired property holds for small numbers.
 A: Let $S_k(x)=\sum_{j=1}^k x^j/j.$ For $0<x<1$ we have $S_k(x)<\sum_{j=1}^{\infty} x^j/j=-\ln (1-x).$ So $T_k=S_k(1/2)<-\ln (1-1/2)=\ln 2.$
So $$0<\sum_{k=1}^{p-2}T_k/(k+1)<(\ln 2)\sum_{k=1}^{p-2}1/(k+1)=(\ln 2)(-1+H(p-1))$$ where $H(p-1)=\sum_{j=1}^{p-1}(1/j)=\gamma+d_{p-1}+\ln (p-1)$
where $\gamma\approx 0.577$ is the Euler-Mascheroni constant and $0<d_{p-1}<1/2(p-1).$ So is there any integer $p\ge 2$ such that $p<(\ln 2)(-1+\gamma+1/2(p-1)+\ln(p-1))$?
A: You might be missing some information in the question because in its current state, there are no possible solutions. The quantity in question is always less than $p$. Here's the reasoning:
Let $S_p = \sum_{k=1}^{p-2}\frac{T_k}{k+1}$ and let $p > 2$
$$T_k = \sum_{j=1}^k {\frac{1}{j2^j}} \leq \sum_{j=1}^k\frac{1}{2^j} < 1$$
$$\implies T_k < 1$$
$$\implies \frac{T_k}{k+1} < \frac{1}{k+1}$$
$$\implies \sum_{k=1}^{p-2}\frac{T_k}{k+1} < \sum_{k=1}^{p-2}\frac{1}{k+1}$$
$$\implies S_p < H_{p-1} - 1 < log_2{(p)} - 1$$
The last step comes from the upper bound of harmonic series. Hence $0<S_p < log_2(p) - 1 < p.$
