# Sum of reciprocals of the perfect powers

• What is the sum of all of the numbers which can be written as the reciprocal of an integer that is also a power greater than one, excluding the powers of one ?.
• I'm asking about the sum of numbers which can be written as powers. $$\mbox{This sum would start off as}\quad \frac{1}{4} + \frac{1}{8} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \cdots$$

Note that I didn't count $1/16$ twice, as we are only adding numbers which can be written in that form.

-I can see how this alternative sum would have probably been easier to handle, since we can see that it equates to the sum $\zeta\left(s\right) - 1$ for all $s$ larger than one. - I know that the first sum must converge, because I have worked out that the sum of numbers which can be written in the form $1/\left(a^{b} - 1\right)$ converges to $1$, where $a,b > 1$.

Is there any way of evaluating the first sum ?. Also, does the second sum converge as well and how can we evaluate it ?.

I have made a program $\left(~\texttt{python}~\right)$ which carries out partial sums for these sorts of series. I'll attach the source code if anyone wants it.

• the first sum should have powers only once, my apologies if that's unclear Oct 20, 2016 at 13:16
• The partial sum for limit $n=10^9$ is ? s=0;for(j=1,10^9,if(ispower(j)>0,s=s+1.0/j));print(s) 0.8744322328535789186668554166 ? (Calculated with PARI/GP) Oct 20, 2016 at 13:18
• yeah, i have tried it using the aforementioned program. Although, mine can only realistically do it to the order 10^5 terms... Oct 20, 2016 at 13:20
• @gebra Take a bit-set to sieve perfect powers first. A one GB array takes you to $2^{33}$, sieving doesn't take long. Sum "downwards", starting from the reciprocals of the largest powers to reduce floating point rounding effects. Oct 20, 2016 at 15:04
• the problem with sieving is you cant run it once and output values which will approach the limit, you'd have to choose the limit first Oct 22, 2016 at 10:45

The sum of the reciprocals of the perfect powers should be $$\sum_{k=2}^{\infty}\sum_{j=2}^{\infty}\frac{(-\mu(k))}{j^k}= \sum_{k=2}^{\infty}\mu(k)\left(1-\zeta(k)\right) \approx 0.874464$$ Where $\mu$ is the Mobius function. See the OEIS sequence A001597.

In order to verify the above formula, it suffices to show that in that double sum each perfect power is counted once. Now $n>1$ is a perfect power iff its prime factorization $p_1^{a_1}\cdots p_{r}^{a_r}$ satisfies $d:=\gcd(a_1,\dots, a_r)>1$.

In the above double sum, the denominator $j^k$ is equal to $n$ iff $k>1$ is a divisor of $d$ and $j=n^{1/k}$. Hence the fraction $1/n$ is counted the following number of times $$\sum_{k\mid d,k>1}(-\mu(k))=0+\mu(1)=1$$ and we are done.

• Why isn't it just the absolute value of $\mu(k)$? Oct 20, 2016 at 18:24
• @Turambar Consider the power $1/2^6$. In the sum with $-\mu(k)$ we will have $1/8^2+1/4^3-1/2^6=1/2^6$. With the the modulus we get three copies of it: $1/8^2+1/4^3+1/2^6=3/2^6$. Oct 20, 2016 at 18:54
• Oh, I misread. I confused $j$ and $k$. In that case, I think it ought to be the absolute value of $\mu(j)$, as you don't want to consider the powers of $1/4$, $1/8$, etc. Or do these work out to be the same? I think this points to the necessity of showing the derivation. Oct 20, 2016 at 19:08
• This expression can be quickly and profitably simplified to $\sum_{k=2}^\infty \mu(k)\big(1-\zeta(k)\big)$; this makes it easy to calculate a huge number of decimal places of the value, since people have worked hard to find fast algorithms for calculating $\zeta(k)$, and the convergence of the infinite sum is awesome since the $k$th term is asymptotic to $2^{-k}$. Oct 20, 2016 at 20:00
• @Turambar I edited my answer with a proof of the formula. Oct 20, 2016 at 20:05

\begin{align} \sum_{n=2}^{\infty} \left [\zeta(n)-1 \right ] &=\sum_{n=2}^{\infty} \sum_{k=2}^{\infty} \frac1{k^n}\\ &= \sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac1{k^n}\\ &= \sum_{k=2}^{\infty} \left ( \frac{k}{k-1} - \frac{k+1}{k}\right ) \\ &= \sum_{k=2}^{\infty} \frac1{k (k-1) }\\ &= 1\end{align}

NB

$$\sum_{n=2}^{\infty} \frac1{k^n} = \sum_{n=0}^{\infty} \frac1{k^n} - 1 - \frac1{k} = \frac{k}{k-1} - \frac{k+1}{k}$$

• :/ This isn't necessarily a bad answer....? Oct 20, 2016 at 14:13
• Note that this answer does count $\frac 1 {16}$ twice, etc. This may explain the difference from Robert Z's answer. Oct 20, 2016 at 17:09
• Hey, this was meant to to be a quick and dirty answer to one of the questions posed above. It was not meant to answer the other question, which apparently was answered by RobertZ. Oct 20, 2016 at 20:49

Your sum contains the sum of all reciprocals of non-unit natural number powers of primes, that is, every number of the form $\frac{1}{p^k}$ for prime $p$, and $k\in \{2,3,...\}=\mathbb{N}\setminus\{1\}$. Since each number of this form is uniquely determined by the choice of $p$ and $k$, we obtain

$\sum_{p}\sum_{i=2}^{\infty}\dfrac{1}{p^i}=\sum_p\dfrac{1}{p(p-1)}$ where $p$ is every prime.

Since $\sum_{i=1}^{\infty}\dfrac{1}{p^i}=\dfrac{1}{p-1}$ (this follows from some basic number theory; an easy way to see it is to consider the repeating decimal, $0.111...$ in base $p$), yet we have removed the leading term, $\dfrac{1}{p}$, hence

$\sum_{p}\sum_{i=2}^{\infty}\dfrac{1}{p^i}=\dfrac{1}{p-1}-\dfrac{1}{p}=\dfrac{p}{p(p-1)}-\dfrac{p-1}{p(p-1)}=\dfrac{p-(p-1)}{p(p-1)}=\dfrac{1}{p(p-1)}$

As for the case at hand, we are no longer concentrating on primes, but the prior counting result holds for non-prime numbers, $n$, the problem now is: how to avoid double counting numbers such as 4^2=4^2=16? Again the solution is to pick on primes, the prime here is $2$, thus we will count that as the $2^4$ and this tactic, in fact, handles all powers of $4$. But that doesn't solve the issue raised by numbers such as $36=2^2\cdot 3^2$. Here we have distinct primes that are part of the factorization, in this case, we can count this as $6^2$. We also have to deal with cases such as

$4^2\cdot 3^2=2^4\cdot 3^2=12^2=144$

For this, we consider it as $12^2$. By now the pattern is maybe emerging: We want take our sum of (non-unit) reciprocal powers over every number whose exponents of its prime factorization have gcd of 1. That is, We are interested in numbers:

$b=p_1^{k_1}\cdot p_2^{k_2}\cdot ...\cdot p_f^{k_f}$

for which $\gcd(k_1,...,k_f)=1$. The reason why this is the criterion we require is that we seek to eliminate recounts of numbers we already have, and any number which has the above $\gcd(k_1,...,k_f)=r\neq 1$, then we can take the $r$th root of that number, and obtain a number which does have $\gcd(k_1,...,k_f)=1$, hence we have a unique way of representing every number we care about as a power of a number in the above form with $\gcd(k_1,...,k_f)=1$.

Thus our sum is, in terms of the above notation:

$\sum_b\sum_{i=2}^{\infty}\dfrac{1}{b^i}=\sum_b\dfrac{1}{b(b-1)}$.

I am not sure if there is a nice way of characterizing the relevant numbers $b$ or not, but provided a suitably nice characterization can be found this may prove useful.

• 6^2 is a power, so yes. But still, i'd like to see where that solution comes from Oct 20, 2016 at 13:17
• $\sum_p 1/p$ is divergent
– mrf
Oct 20, 2016 at 13:22
• I made a few errors in coming up with that, working on a fix. Oct 20, 2016 at 13:24
• The inner sum should start at $i=2$, otherwise the sum clearly diverges as mrf notes. Oct 20, 2016 at 13:31
• @gebra Inclusion-exclusion principle. Say $n > 1$ is a perfect power. You can write it $n = b^k$ with maximal $k$. Then all ways to write $n$ as a (proper) power are $n = (b^{k/d})^d$ with $d$ dividing $k$ and $d > 1$. You want $\frac{1}{n}$ to be counted only once. Noting $\sum\limits_{d\mid k} \mu(d) = 0$ since $k > 1$, it follows that $$1 = \mu(1) = \sum_{\substack{d \mid k \\ d > 1}} (-\mu(d)).$$ Oct 20, 2016 at 14:54