TITLE: Proving $\lim_{n\to\infty}\left(n-\sum_{k=2}^{n}\frac{1}{\sum_{i=1}^{\infty}\frac{1}{i^k}}\right)=1+\sum_{p\in P}\frac{1}{p\left(p-1\right)}$?
QUESTION [8 upvotes]: $$\lim_{n\to\infty}\left(n-\sum_{k=2}^{n}\frac{1}{\sum_{i=1}^{\infty}\frac{1}{i^k}}\right)=1+\sum_{p\in P}\frac{1}{p\left(p-1\right)}$$
$P$ is primes.
Interesting question ran across while tutoring. Not sure how to solve it, or even what class it's from.

REPLY [8 votes]: If I calculated correctly, it seems that the answer is wrong. Indeed,
\begin{align*}
n-\sum_{k=2}^{n}\frac{1}{\zeta(k)}
&= n-\sum_{k=2}^{n}\sum_{m=1}^{\infty} \frac{\mu(m)}{m^k}
 = 1-\sum_{k=2}^{n}\sum_{m=2}^{\infty} \frac{\mu(m)}{m^k}
 = 1-\sum_{m=2}^{\infty} \mu(m) \sum_{k=2}^{n}\frac{1}{m^k},
\end{align*}
and taking $n\to\infty$ we obtain
\begin{align*}
\lim_{n\to\infty} \left( n-\sum_{k=2}^{n}\frac{1}{\zeta(k)} \right)
&= 1-\sum_{m=2}^{\infty} \mu(m) \sum_{k=2}^{\infty}\frac{1}{m^k}
 = 1-\sum_{m=2}^{\infty} \frac{\mu(m)}{m(m-1)} \\
&= 1 + \sum_{p\in\Bbb{P}} \frac{1}{p(p-1)} - \sum_{p,q\in\Bbb{P}} \frac{1}{pq(pq-1)} + \cdots.
\end{align*}
Numerical calculations with aid of Mathematica also suggests that
$$ \lim_{n\to\infty} \left( n-\sum_{k=2}^{n}\frac{1}{\zeta(k)} \right) \approx 1.70521, $$
while
$$ 1 + \sum_{p\in\Bbb{P}} \frac{1}{p(p-1)} \approx 1.77316. $$