# Sum of series $\sum_{k=1}^\infty{-\frac{1}{2^k-1}}$

What is the sum of the series $\displaystyle\sum_{k=1}^\infty{\frac{-1}{2^k-1}}$?

Also, more generally, can we find $\displaystyle\sum_{k=1}^\infty{\frac{-1}{c^k-1}}$ for some $c$?

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Do you mean $(-1)^k$? If not, why include the minus sign? – joriki Apr 5 '11 at 5:41
@joriki: No. I guess I just did it for correctness. – Matt Groff Apr 5 '11 at 14:05
I don't understand -- correctness of what? In what sense would the simpler sum without the minus sign not have been correct? – joriki Apr 5 '11 at 14:07
@joriki: I guess it's pretty pointless to include the minus. Unfortunately, the answers here correspond with it included. I'll remember to simplify things more for future reference. Thanks for the tip. – Matt Groff Apr 5 '11 at 15:47

Ignoring the negative sign, you're looking for

$${1 \over 2^1 - 1} + {1 \over 2^2 - 1} + {1 \over 2^3 - 1} + \cdots$$

but we have

$${1 \over 2^{jk} - 1} = {1 \over 2^{jk}} + {1 \over 2^{2jk}} + {1 \over 2^{3jk}} + \cdots.$$

If we rewrite the first expression using the second one, then your sum is

$$\left( {1 \over 2^1} + {1 \over 2^2} + {1 \over 2^3} + \cdots \right) + \left( {1 \over 2^2} + {1 \over 2^4} + {1 \over 2^6} + \cdots \right) + \left( {1 \over 2^3} + {1 \over 2^6} + {1 \over 2^9} + \cdots \right) + \cdots$$

and $1/2^r$ appears $\tau(r)$ times, where $\tau(r)$ is the number of divisors of $r$. Therefore your sum is

$$\sum_{r \ge 1} \tau(r) 2^{-r}$$

and this should allow you to compute it to any desired level of numerical accuracy fairly quickly. For example,

$$\sum_{r=1}^{40} \tau(r) 2^{-r} = {27602812537 \over 17179869184} = 1.606695152$$

and the error here is less than $\sum_{r \ge 41} r 2^{-r} = 84/2^{41} < 4 \times 10^{-11}$ since $\tau(r) < r$ for all $r \ge 3$.

Of course this method works when $2$ is replaced by any constant $c > 1$.

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Very clever use of the geometric series (which I lectured about in my Calc II class yesterday)! For those algorithmically minded folks in the audience, what clever ways are there to find $\tau(n)$? We can muck about with properties given a prime factorization, but would that save us much time/computation? I'd be happy to break this off into my own question, if desired. – user4689 Apr 5 '11 at 20:25
I'm not really a number theorist. But computing $\tau(n)$ in general would be at least as hard as primality testing, since $n$ is prime if and only if $\tau(n) = 2$. – Michael Lugo Apr 5 '11 at 21:53
Via Wikipedia's take on the subject, I played around with computing $\tau(n)$ in Sage on my poor little laptop. Things seemed to be faster if we went with $\Pi_1^n (a_i + 1)$ (where $n = \Pi_1^n p_i^{a_i}$ is the prime factorization of $n$) than the naive approach, but that may just be my computer and poor skills. – user4689 Apr 5 '11 at 23:20
What's the "naive approach" here? To be honest, I would have called the approach you used the naive approach. – Michael Lugo Apr 5 '11 at 23:29
Fair enough. I had in mind just trial division. In (unoptimized) Python code, something like sum(1 for i in xrange(1, n + 1) if n % i == 0). – user4689 Apr 6 '11 at 3:25

Wolfram Alpha gives an explicit answer for c=2, approximately -1.6067, and an explicit mess for arbitrary c.

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$$\sum_{k=1}^\infty c^{-k}\left(1-\left(\frac{1}{c} \right)^k \right)^{-1} ;c>1$$

$$\sum_{k=1}^\infty c^{-k} +\sum_{k=1}^\infty c^{-2k} +\sum_{k=1}^\infty c^{-3k} \cdots = \sum_{k=1}^\infty \sum_{n=1}^\infty c^{-n k}$$

In the double summation written above note that ($s=c^{-1}$) , $s^l$ appears as many times as there are solutions to $nk=l$ where $n,k,l\in \mathbb{N}$. Let $f(l)$ be the number of solutions, then

$$\sum_{k=1}^\infty \sum_{n=1}^\infty s^{n k} = \sum_{l=1}^\infty f(l) s^l ; s<1$$

The coefficient $f(l)$ is studied in more detail in number theory as it appears in the square of the zeta function. $\zeta (s)^2 = \sum_{n=1}^\infty \frac{f(n)}{n^s}$

The $f(n)$ is called the divisor function. So one way will be to write the series as a power series with the coefficients $f(n)$.

Other option is to use Computer algebra, as Ross answered. Mathematica gives a closed form

$$\frac{-\text{Log}[-1+c]-\text{Log}\left[\frac{1}{c}\right]-\text{QPolyGamma}\left[0,1,\frac{1}{c}\right]}{\text{Log}[c]}$$

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It is the Erdős-Borwein Constant. (modulo a minus sign)

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