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Find sum of$\sum_{n=1}^\infty {\frac {f(n)} {n(n+1)}}$ where $f(n)$ is the seriesnumber of $1$s in $n$'s binary expansion

We are given the series $\sum_{n=1}^\infty {\frac {f(n)} {n(n+1)}}$, where $f(n)$ is such a function that it equals the sum of 1's in the binary representation of n.

I'm obliged to find the sum of the series. First, I decided to explicitly prove the convergence of the series. Use for that converging harmonic series,

$\sum {\frac 1 {n^2}}$.
$\lim_{n \to \infty} \frac {f(n)* n^2} {n \cdot (n+1)} = \lim_{n \to \infty} f(n)$

Where $f(n)$ could be represented as $\sum_{i=0}^{\log_2 n} m_i$, if we define n as $\sum_{i=0}^{\log_2 n} m_i \cdot 2^i$ and $m_i$ is able to take values of 0 or 1.

I can get it intuitively that $f(n)$ doesn't yield $\infty$, when $n \to \infty$. A graph of the function would behave like wave, growing and then returning to 1. I even wrote a script to get some sense of the data, which outputted:

For p= 10 series= 1.0873015873015872
For p= 100 series= 1.3398423025935298
For p= 1000 series= 1.3801052489558094
For p= 100000 series= 1.3861981879674006

This result looks for me like $ln(4)$, but thatit doesn't help much, since the solution is supposed to be purely analytical. How docan I find a sum?

We are given the series $\sum_{n=1}^\infty {\frac {f(n)} {n(n+1)}}$, where $f(n)$ is such a function that it equals the sum of 1's in the binary representation of n.

I'm obliged to find the sum of the series. First, I decided to explicitly prove the convergence of the series. Use for that converging harmonic series,

$\sum {\frac 1 {n^2}}$.
$\lim_{n \to \infty} \frac {f(n)* n^2} {n \cdot (n+1)} = \lim_{n \to \infty} f(n)$

Where $f(n)$ could be represented as $\sum_{i=0}^{\log_2 n} m_i$, if we define n as $\sum_{i=0}^{\log_2 n} m_i \cdot 2^i$ and $m_i$ is able to take values of 0 or 1.

I can get it intuitively that $f(n)$ doesn't yield $\infty$, when $n \to \infty$. A graph of the function would behave like wave, growing and then returning to 1. I even wrote a script to get some sense of the data, which outputted:

For p= 10 series= 1.0873015873015872
For p= 100 series= 1.3398423025935298
For p= 1000 series= 1.3801052489558094
For p= 100000 series= 1.3861981879674006

This result looks for me like $ln(4)$, but that doesn't help much, since the solution is supposed to be purely analytical. How do I find a sum?

We are given the series $\sum_{n=1}^\infty {\frac {f(n)} {n(n+1)}}$, where $f(n)$ is such a function that it equals the sum of 1's in the binary representation of n.

I'm obliged to find the sum of the series. First, I decided to explicitly prove the convergence of the series. Use for that converging harmonic series,

$\sum {\frac 1 {n^2}}$.
$\lim_{n \to \infty} \frac {f(n)* n^2} {n \cdot (n+1)} = \lim_{n \to \infty} f(n)$

Where $f(n)$ could be represented as $\sum_{i=0}^{\log_2 n} m_i$, if we define n as $\sum_{i=0}^{\log_2 n} m_i \cdot 2^i$ and $m_i$ is able to take values of 0 or 1.

I can get it intuitively that $f(n)$ doesn't yield $\infty$, when $n \to \infty$. A graph of the function would behave like wave, growing and then returning to 1. I wrote a script to get some sense of the data, which outputted:

For p= 10 series= 1.0873015873015872
For p= 100 series= 1.3398423025935298
For p= 1000 series= 1.3801052489558094
For p= 100000 series= 1.3861981879674006

This result looks for me like $ln(4)$, but it doesn't help much, since the solution is supposed to be purely analytical. How can I find a sum?

find Find sum of the series

We are given the series $\sum_{n=1}^\infty {\frac {f(n)} {n(n+1)}}$, where $f(n)$ is such a function that it equals to the sum of 1's in the binary representation of n.
I'm

I'm obliged to find the sum of the series.
first First, I decided to explicitly prove the convergence of the series. Use Use for that converging harmonicharmonic series  ,

$\sum {\frac 1 {n^2}}$.
$\lim_{n \to \infty} \frac {f(n)* n^2} {n \cdot (n+1)} = \lim_{n \to \infty} f(n)$
Where

Where $f(n)$ could be represented as $\sum_{i=0}^{\log_2 n} m_i$, if we define n as $\sum_{i=0}^{\log_2 n} m_i \cdot 2^i$ and $m_i$ ableis able to take values of 0 or 1.

I can get it intuitively that $f(n)$ doesn't yield $\infty$, when $n \to \infty$. GraphA graph of the function would behave like wave, growing and then returning to 1. I even wrote a script to get some sense of the data, which outputted:

For p= 10 series= 1.0873015873015872
For p= 100 series= 1.3398423025935298
For p= 1000 series= 1.3801052489558094
For p= 100000 series= 1.3861981879674006

This result looks for me like $ln(4)$, but that doesn't help much, since the solution is supposed to be purely analytical. So any help with finding How do I find a sum would be appreciated.?

find sum of the series

We are given the series $\sum_{n=1}^\infty {\frac {f(n)} {n(n+1)}}$, where $f(n)$ is such a function that it equals to the sum of 1's in the binary representation of n.
I'm obliged to find the sum of the series.
first, I decided to explicitly prove the convergence of series. Use for that converging harmonic series  $\sum {\frac 1 {n^2}}$.
$\lim_{n \to \infty} \frac {f(n)* n^2} {n \cdot (n+1)} = \lim_{n \to \infty} f(n)$
Where $f(n)$ could be represented as $\sum_{i=0}^{\log_2 n} m_i$, if we define n as $\sum_{i=0}^{\log_2 n} m_i \cdot 2^i$ and $m_i$ able to take values of 0 or 1.

I can get it intuitively that $f(n)$ doesn't yield $\infty$, when $n \to \infty$. Graph of the function would behave like wave, growing and then returning to 1. I even wrote a script to get some sense of the data, which outputted:

For p= 10 series= 1.0873015873015872
For p= 100 series= 1.3398423025935298
For p= 1000 series= 1.3801052489558094
For p= 100000 series= 1.3861981879674006

This result looks for me like $ln(4)$ but that doesn't help much, since solution supposed to be purely analytical. So any help with finding a sum would be appreciated.

Find sum of the series

We are given the series $\sum_{n=1}^\infty {\frac {f(n)} {n(n+1)}}$, where $f(n)$ is such a function that it equals the sum of 1's in the binary representation of n.

I'm obliged to find the sum of the series. First, I decided to explicitly prove the convergence of the series. Use for that converging harmonic series,

$\sum {\frac 1 {n^2}}$.
$\lim_{n \to \infty} \frac {f(n)* n^2} {n \cdot (n+1)} = \lim_{n \to \infty} f(n)$

Where $f(n)$ could be represented as $\sum_{i=0}^{\log_2 n} m_i$, if we define n as $\sum_{i=0}^{\log_2 n} m_i \cdot 2^i$ and $m_i$ is able to take values of 0 or 1.

I can get it intuitively that $f(n)$ doesn't yield $\infty$, when $n \to \infty$. A graph of the function would behave like wave, growing and then returning to 1. I even wrote a script to get some sense of the data, which outputted:

For p= 10 series= 1.0873015873015872
For p= 100 series= 1.3398423025935298
For p= 1000 series= 1.3801052489558094
For p= 100000 series= 1.3861981879674006

This result looks for me like $ln(4)$, but that doesn't help much, since the solution is supposed to be purely analytical. How do I find a sum?

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