How find this sum $\sum_{n=1}^{\infty}n\sum_{k=2^{n-1}}^{2^n-1}\frac{1}{k(2k+1)(2k+2)}$ Find the sum
$$\sum_{n=1}^{\infty}n\sum_{k=2^{n - 1}}^{2^{n}\ -\ 1}\dfrac{1}{k(2k+1)(2k+2)}$$
My try:
note
$$\dfrac{1}{k(2k+1)(2k+2)}=\dfrac{2}{(2k)(2k+1)(2k+2)}=\left(\dfrac{1}{(2k)(2k+1)}-\dfrac{1}{(2k+1)(2k+2)}\right)$$
Then I can't 

Are you really sure about the outer $n$? If not, Wolfram Alpha gives a sum that telescopes nicely
$$\sum_{k=2^{n-1}}^{2^n-1} \frac{1}{k (2 k+1) (2 k+2)}=-2^{-n-1}-\psi^{(0)}(2^{n-1})+\psi^{(0)}(2^n)-\psi^{(0)}\left(\frac{1}{2}+2^n\right) +\psi^{(0)}\left(\frac{1}{2}+2^{n-1}\right)$$
Update:
It is not hard to prove user64494's result manually
$$S_2=\sum_{n=1}^{\infty}\sum_{k=2^{n-1}}^{2^n-1}\frac{1}{k(2k+1)(2k+2)}=$$
$$\sum_{n=1}^{\infty}\frac{1}{k(2k+1)(2k+2)}=$$
$$\sum_{n=1}^{\infty}2\left(\frac{1}{2k}-\frac{1}{2k+1}\right)-\left( \frac{1}{2k}-\frac{1}{2k+2}\right)=$$
$$2\left(\sum_{n=1}^{\infty}\frac{1}{2k}-\frac{1}{2k+1}\right)-\frac12\sum_{n=1}^{\infty}\frac1k-\frac{1}{k+1}=$$
Since $\ln{2}=1-\frac12+\frac13-\cdots$, we have
$$1-\ln{2}=\sum_{n=1}^{\infty}\frac{1}{2k}-\frac{1}{2k+1}$$
And the second sum is a nicely telescopic one that equals $1$. Therefore
$$S_2=2(1-\ln{2})-\frac12=\frac32-2\ln{2}$$
But this is unfortunately not what you have asked for yet. The real sum you are asking for doesn´t get a closed form from either WolframAlpha or Maple, so I don´t think there is an easy way to get it, at least for me.
