Find sum of $\sum^{\infty}_{n=1}\frac{f(n)}{n(n+1)}$, $f(n)$ - the number of ones in the binary representation of the number $n$.

Question

Here is task:

$$\sum_{n=1}^{\infty}\frac{f(n)}{n(n+1)}$$

Where $f(n)$ - the number of ones in the binary representation of the number n.

I tried separate $a_n$ to $\frac{f(n)}{n} - \frac{f(n)}{n+1}$ then we have $$ \sum_{n=1}^{\infty}\frac{f(n)}{n(n+1)} = 1 + \sum_{n=2}^{\infty}\frac{f(n) - f(n - 1)}{n} $$ $$ f(n) = \begin{cases}1&n=1\\f(z)& n = 2z\\f(z) + 1&n = 2z + 1\end{cases} $$

Then we have, if n is odd in second sum: $$ f(2z + 1) - f(2z)= f(z) + 1 - f(z) = 1 $$

Can't understand what I should to do next.

I believe it equals $\ln(4)$. See https://oeis.org/A000120 – Mar 05 '23 at 21:49 — , Mar 05 '23 at 21:49

Sangchul Lee · Accepted Answer · 2023-03-07T21:00:00.823

4

We have

$$ f(n) = \sum_{k=0}^{\infty} \mathbf{1}\{\lfloor n / 2^k \rfloor \text{ is odd}\}, $$

where $\mathbf{1}\{\ldots\}$ stands for the indicator function. Plugging this into OP's sum and interchanging the order of summation (which is allowed by Fubini–Tonelli Theorem),

\begin{align*} \sum_{n=1}^{\infty} \frac{f(n)}{n(n+1)} &= \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \sum_{k=0}^{\infty} \mathbf{1}\{\lfloor n / 2^k \rfloor \text{ is odd}\} \\ &= \sum_{k=0}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \mathbf{1}\{\lfloor n / 2^k \rfloor \text{ is odd}\}. \end{align*}

Now we focus on the inner sum that runs over $n = 1, 2, \ldots$. Since the indicator function factor survives with value $1$ precisely when $\lfloor n / 2^k \rfloor$ is an odd number, we may represent those $n$ by the form

$$ n = (2q-1)2^k + r, $$

where $q = 1, 2, 3, \ldots$ and $r = 0, 1, \ldots, 2^k - 1$. So,

\begin{align*} &\sum_{n=1}^{\infty} \frac{1}{n(n+1)} \mathbf{1}\{\lfloor n / 2^k \rfloor \text{ is odd}\} \\ &= \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) \mathbf{1}\{\lfloor n / 2^k \rfloor \text{ is odd}\} \\ &= \sum_{q=1}^{\infty} \sum_{r=0}^{2^k - 1} \left( \frac{1}{(2q-1)2^k + r} - \frac{1}{(2q-1)2^k + r + 1} \right) \\ &= \sum_{q=1}^{\infty} \left( \frac{1}{(2q-1)2^k} - \frac{1}{(2q-1)2^k + 2^k} \right) \\ &= \frac{1}{2^k} \sum_{q=1}^{\infty} \left( \frac{1}{2q-1} - \frac{1}{2q} \right). \end{align*}

In the third step, we utilized the fact that the inner sum is telescoping. However, the last sum is the famous alternating harmonic series, which converges to $\log 2$. Therefore,

\begin{align*} \sum_{n=1}^{\infty} \frac{f(n)}{n(n+1)} = \sum_{k=0}^{\infty} \frac{\log 2}{2^k} = 2 \log 2. \end{align*}

edited Mar 07 '23 at 21:00

answered Mar 06 '23 at 03:52

Sangchul Lee

181,930

Can't understand this rows $$\begin{align} &\sum_{k=0}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \mathbf{1}{\lfloor n / 2^k \rfloor \text{ is odd}} \ &= \sum_{k=0}^{\infty} \sum_{q=1}^{\infty} \left( \frac{1}{(2q-1)2^k} - \frac{1}{(2q)2^k} \right) \ &= \sum_{k=0}^{\infty} \frac{\log 2}{2^k} \ \end{align}$$ – replikeit Mar 06 '23 at 18:22
@replikeit, Group the inner sum according to the value of $q = \lfloor n/2^k\rfloor = 1,3,5,7,\ldots$, and see how the sum over each group simplifies :) – Sangchul Lee Mar 06 '23 at 18:24
Okay, but where log came from??? – replikeit Mar 06 '23 at 18:58
@replikeit, Check the alternating harmonic series: $$1-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\cdots=\log 2. $$ – Sangchul Lee Mar 06 '23 at 19:04
Thank you a lot!! – replikeit Mar 06 '23 at 20:59
I tried to group but have not success it, didn't understand why $2^k$ is here, I think I missed something. Sorry for questions, this is the most hardest sum for me, before I have not experience with tasks like this. Could you answer this step more openly? – replikeit Mar 06 '23 at 23:53
@replikeit, No worries. I added more details. – Sangchul Lee Mar 07 '23 at 03:36
Thank you, you are my saver!!!!! – replikeit Mar 07 '23 at 09:37
We getting $n$ only when this $n\mod{(2^k)}$ is odd. Then we have $n = (2q - 1)2^k + r$, where $r = 0, 1, 2, 3 \dots, 2^k - 1$. But if we get $n_0 = (2q - 1)2^k + 0.$ We have that, $n_0\mod{2^k}=0$. – replikeit Mar 07 '23 at 15:22
Let us continue this discussion in chat. – replikeit Mar 07 '23 at 20:58

score 1 · Answer 2 · answered Mar 07 '23 at 19:06

I found another solution. This is not elegantly like answer before, but more simply.

\begin{align*} \sum\limits_{n=1}^\infty \frac{f(n)}{n(n+1)} &= \frac12 + \sum\limits_{n=1}^\infty \frac{f(2n)}{2n(2n+1)} + \sum\limits_{n=1}^\infty \frac{f(2n+1)}{(2n+1)(2n+2)} \\ &= \frac12 + \sum\limits_{n=1}^\infty \frac{f(n)}{2n(2n+1)} + \sum\limits_{n=1}^\infty \frac{f(n)+1}{(2n+1)(2n+2)} \\ &= \frac12 + \sum\limits_{n=1}^\infty f(n) \left[ \frac{1}{2n(2n+1)} + \frac{1}{(2n+1)(2n+2)} \right] + \sum\limits_{n=1}^\infty \frac{1}{(2n+1)(2n+2)} \\ &= \frac12 + \frac12 \sum\limits_{n=1}^\infty \frac{f(n)}{n(n+1)} + \sum\limits_{n=1}^\infty \frac{1}{(2n+1)(2n+2)}. \end{align*} Then we have \begin{align*} \sum\limits_{n=1}^\infty \frac{f(n)}{n(n+1)} &= 1 + 2\sum\limits_{n=1}^\infty \frac{1}{(2n+1)(2n+2)} = 1 + 2\sum_{n=3}^\infty \frac{(-1)^{n+1}}{n} \\ &= 2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = 2 \left. \sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n} \right|_{x=1} = 2\ln (1+x) |_{x=1} = 2\ln 2. \end{align*}

Find sum of $\sum^{\infty}_{n=1}\frac{f(n)}{n(n+1)}$, $f(n)$ - the number of ones in the binary representation of the number $n$.

2 Answers2