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.

Solution 1:

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.

Solution 2:

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

NB

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