1
$\begingroup$

Is it possible to find a closed expression for the following sum:

$\sum_{k=1}^{\infty}\frac{1}{a_k^2}$

where $a_k$ is the $k$-th number with its most significant '$1$' at an odd digit place?

A number of this kind is the number $5$, because $5$ in binary is $101_2$, which has its most significant '$1$' at place three. A list of these numbers is given by sequence A053738.

$\endgroup$
4
  • $\begingroup$ That is not $5$ in binary. This is how $5$ is written in binary: $$101_2$$ $\endgroup$
    – Franklin Pezzuti Dyer
    Commented Jul 10, 2017 at 20:53
  • $\begingroup$ I'm not understanding... when we say something like $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ we usually mean that $k$ is an index variable for the summation and the summation refers to $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\dots$... here you talk about "where $k$ is a number with its most significant $1$ at an odd digit place"... do you mean to do instead the sum $\sum\limits_{k\in K}\frac{1}{k^2}$ where $K$ is the set of numbers with most significant $1$'s at an odd digit place (in binary)? $\endgroup$
    – JMoravitz
    Commented Jul 10, 2017 at 20:53
  • $\begingroup$ Yes, I mean the k-th number in the sequence A053738. $\endgroup$
    – gilianzz
    Commented Jul 10, 2017 at 20:55
  • $\begingroup$ Alternatively, write it as $\sum\limits_{k=1}^\infty \frac{1}{a_k^2}$ where $a_k$ is the $k$'th number with its most significant $1$ at an odd digit place... but don't use $k$ to represent multiple different things in the same sentence $\endgroup$
    – JMoravitz
    Commented Jul 10, 2017 at 20:56

1 Answer 1

Reset to default
2
$\begingroup$

Your sequence $a_k$ consists of the numbers $2^{2j}$ to $2^{2j+1}-1$, for nonnegative integers $j$. Thus your sum is

$$ \sum_{j=0}^\infty \sum_{i=2^{2j}}^{2^{2j+1}-1} \frac{1}{i^2} = \sum_{j=0}^\infty \left(\Psi(1,2^{2j}) -\Psi \left( 1,2^{2j+1} \right) \right) = \sum_{k=0}^\infty (-1)^k \Psi(1, 2^k)$$

in Maple's notation. I don't know of a closed form for the last sum. Numerically it is approximately $1.1939512459529255006$. The Inverse Symbolic Calculator returns nothing.

$\endgroup$
2
  • $\begingroup$ What is the psi-function doing? $\endgroup$
    – gilianzz
    Commented Jul 10, 2017 at 21:18
  • 1
    $\begingroup$ $\Psi(1,x) = \dfrac{d}{dx} \Psi(x)$. If $x$ is a positive integer, $\Psi(1,x) = \sum_{j=x}^\infty 1/j^2$. $\endgroup$
    – Robert Israel
    Commented Jul 10, 2017 at 21:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .