How find this sum $\sum_{n=1}^{\infty}n\sum_{k=2^{n-1}}^{2^n-1}\frac{1}{k(2k+1)(2k+2)}$

Question

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

Answer 1

Throughout, let $H_k\equiv \sum_{j=1}^k \frac{1}{j}$ and let $\gamma$ be the Euler-Mascheroni constant.

First, note \begin{eqnarray} \sum_{n=1}^{\infty} n \sum_{k=2^{n-1}}^{2^n-1} \frac{1}{k(2k+1)(2k+2)} &=& \sum_{j=0}^{\infty} \sum_{k=2^j}^{\infty} \frac{1}{k(2k+1)(2k+2)}. \end{eqnarray}

Mathematica says \begin{eqnarray} \sum_{k=2^j}^{\infty} \frac{1}{k(2k+1)(2k+2)} = -\frac{1}{2}\psi(2^j) + \psi(2^j+\frac{1}{2}) - \frac{1}{2} \psi(2^j+1) \end{eqnarray} where $\psi$ is the digamma function. To compute this without as much help from softare, use the original poster's expression for the summand and the fact that $$ \psi(z+1) = -\gamma + \sum_{n=1}^{\infty} \frac{z}{n(n+z)} $$ for any $z\in\mathbb{C}$ as long as $z\ne -1,-2,-3,\ldots$.

Using explicit formulas (see Wikipedia) for $\psi(m)$ and $\psi(m+\frac{1}{2})$ where $m$ is an integer, \begin{eqnarray} -\frac{1}{2}\psi(2^j) + \psi(2^j+\frac{1}{2}) - \frac{1}{2} \psi(2^j+1) &=& -\frac{1}{2}H_{2^j-1} - \frac{1}{2} H_{2^j} - 2 \log 2 + \sum_{k=1}^{2^j} \frac{2}{2k-1} \\ &=& -\frac{1}{2}H_{2^j-1} - \frac{1}{2} H_{2^j} - 2 \log 2 + 2(H_{2^{j+1}} - \frac{1}{2} H_{2^j}) \\ &=& -\frac{1}{2}H_{2^j-1} - \frac{3}{2} H_{2^j} - 2 \log 2 + 2 H_{2^{j+1}} \\ &=& -\frac{1}{2}(H_{2^j} -\frac{1}{2^j}) - \frac{3}{2} H_{2^j} - 2 \log 2 + 2 H_{2^{j+1}} \\ &=& \frac{1}{2^{j+1}} - 2 \log 2 + 2 (H_{2^{j+1}} - H_{2^j}). \end{eqnarray} Observe \begin{eqnarray} \sum_{j=0}^m \frac{1}{2^{j+1}} - 2 \log 2 + 2 (H_{2^{j+1}} - H_{2^j}) &=& 1 - 2^{-(m+1)} + 2\left(H_{2^{m+1}}- \log 2^{m+1}\right) - 2H_1. \end{eqnarray} It follows \begin{eqnarray} \sum_{j=0}^{\infty} \sum_{k=2^j}^{\infty} \frac{1}{k(2k+1)(2k+2)} &=& \lim_{m\to\infty} 1 - 2^{-(m+1)} + 2\left(H_{2^{m+1}}- \log 2^{m+1}\right) - 2H_1 \\ &=& 2\gamma - 1. \end{eqnarray}

Answer 2

Not an answer, just a (possibly useless) comment (or commentary? I don't speak English). We have

$$\sum_{n=1}^{\infty}n\sum_{k=2^{n-1}}^{2^n-1}\dfrac{1}{k(2k+1)(2k+2)}=\sum_{k=1}^\infty\dfrac{1}{k(2k+1)(2k+2)}\sum_{\substack{n\geq1\\2^{n-1}\leq k<2^n}}n\,;$$

but given $k\geq1$, then $2^{n-1}\leq k<2^n$ iff $n=\Bigl\lfloor1+\dfrac{\log k}{\log2}\Bigr\rfloor$, so the inner sum is actually equal to this value of $n$, and so the required sum is equal to

$$\sum_{k=1}^\infty\dfrac{1}{k(2k+1)(2k+2)}\cdot\biggl\lfloor1+\dfrac{\log k}{\log2}\biggr\rfloor\,.$$

Answer 3

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.