Sum Of Squared Reciprocals - Mobius Function

We can rewrite this as $\sum_{n=1}^\infty \dfrac{f(n)}{n^2}$ where $f(n)=\begin{cases} 1 && ,\mu(n)=1\\ 0 && ,\mu(n)=0\\ 0 && ,\mu(n)=-1\end{cases}$ . for it to be zero when $\mu$ is, $\mu$ must be one of its factor. For it to be zero when $\mu=-1\to \mu+1=0$ , $\mu+1$ must be its factor. But when $\mu=1$ , the second part will be 2, so we need to divide by 2 to make it one. We find $f(n)=\dfrac{\mu(n)(\mu(n)+1)}{2}=\dfrac{|\mu(n)|}{2}+\dfrac{\mu(n)}{2}$ Hence the summation becomes $\dfrac{1}{2}\left(\sum_{n=1}^\infty\dfrac{|\mu(n)|}{n^2}+\sum_{n=1}^\infty\dfrac{\mu(n)}{n^2}\right)$ Using dirichlet series and dirichlet convolution , $\sum_{n=1}^\infty\dfrac{\mu(n)}{n^s}=\dfrac{1}{\zeta(s)}\\|\mu|*\lambda=I\Longrightarrow \sum_{n=1}^\infty\dfrac{|\mu(n)|}{n^s}=\dfrac{1}{\sum_{n=1}^\infty\dfrac{\lambda(n)}{n^s}}=\dfrac{\zeta(s)}{\zeta(2s)}$

From the riemann zeta function , we know $\zeta(2)=\dfrac{\pi^2}{6},\zeta(4)=\dfrac{\pi^4}{90}$ . Substituting in these values gives $\dfrac{21}{2\pi^2},$ so $A+B+C = 21+2+2 = 25.$

First, rewrite the given sum $s$ . If we add $\mu(n)$ and $\mu^2(n)$ , the terms where $\mu(n)=-1$ cancel out. However the terms where $\mu(n)=1$ we count double. Sums and products over $p_k$ denote sums and products over all primes: $\begin{aligned} 2s &= 2\sum_{\mu(n)=1} \frac{1}{n^2}= \sum_{\mu(n)\neq 0} \frac{\mu^2(n) + \mu(n)}{n^2}&&&\left|\:\: \zeta(k)=\sum_{n\in\mathbb{N}}\frac{1}{n^{k}}=\prod_{p_1}\frac{1}{1-\frac{1}{p_1^k}},\right.&&k\in\mathbb{N},&&k&>1 \end{aligned}$

Both sums on the left converge absolutely, so we may reorder them in any way we want. The first reordering combines all $n$ with the same number of prime factors. The second reordering takes only the first few summands of the first few sums to convert them into a growing product over all primes that resembles the zeta-function above: $\begin{aligned} \sum_{\mu(n)\neq 0} \frac{\mu^2(n)}{n^2}& = 1 + \sum_{p_1}\frac{1}{p_1^2} + \sum_{p_1 < p_2}\frac{1}{p_1^2p_2^2} + \sum_{p_1 < p_2<p_3}\frac{1}{p_1^2p_2^2p_3^2} + \ldots = \prod_{p_1}\left(1 + \frac{1}{p_1^2}\right)=\prod_{p_1}\frac{1 - \frac{1}{p_1^4}}{1 - \frac{1}{p_1^2}}=\frac{\zeta(2)}{\zeta(4)}\\[1em] \sum_{\mu(n)\neq 0} \frac{\mu(n)}{n^2}& = 1 \red{- \sum_{p_1}\frac{1}{p_1^2}} + \sum_{p_1 < p_2}\frac{1}{p_1^2p_2^2} \red{- \sum_{p_1 < p_2<p_3}\frac{1}{p_1^2p_2^2p_3^2}} \pm \ldots = \prod_{p_1}\left(1 - \frac{1}{p_1^2}\right)=\frac{1}{\zeta(2)} \end{aligned}$

Adding both together, the red terms cancel out and we are left with $2s$ . We remember $\zeta(2)=\frac{\pi^2}{6},\quad \zeta(4)=\frac{\pi^4}{90}$ to get $\begin{aligned} 2s &= \frac{\pi^2\cdot 90}{\pi^4\cdot 6} + \frac{6}{\pi^2} = \frac{21}{\pi^2}&&&\Rightarrow &&&&s&=\frac{21}{2\pi^2},&&& A+B+C&=21+2+2=\boxed{25} \end{aligned}$