350 followers problem

I've already solved it here , the solution goes as :

We have by Drichlet Convolution $(\phi*I)(n) = n$

$I(n)$ Denotes the Identity Function .

$\mathbf{Proof }$

$(\displaystyle\phi*I)(n) = \sum_{d|n} \phi(d)I(\frac{n}{d}) = \sum_{d|n}\phi(d) = n$

$\text{Of course it is a well known identity that }\sum_{d|n}\phi(d) = n$

Next we have :

$\displaystyle\sum_{n\in\mathbb{N}} \frac{(f*g)(n)}{n^s} = \sum_{n\in\mathbb{N}} \frac{f(n)}{n^s} \sum_{n\in\mathbb{N}} \frac{g(n)}{n^s}$

$\mathbf{Proof}$

We may write the summation as :

$\displaystyle\sum_{n\in\mathbb{N}} \frac{(f*g)(n)}{n^s}=\sum_{n\in\mathbb{N}} \sum_{n=d_1d_2} \frac{f(d_1)}{{d_1}^s}\frac{f(d_2)}{{d_2}^s}$

Note that every ordered pair $(d_1,d_2)$ occurs exactly once namely $n=d_1d_2$

So the above summation can be rewritten as :

$\displaystyle \sum_{n\in\mathbb{N}} \frac{(f*g)(n)}{n^s}=\sum_{d_1\in\mathbb{N}} \frac{f(d_1)}{{d_1}^s} \sum_{d_1\in\mathbb{N}} \frac{f(d_2)}{{d_2}^s}$

Thus we have

$\displaystyle \sum_{n\in\mathbb{N}} \frac{(f*g)(n)}{n^s} = \sum_{n\in\mathbb{N}} \frac{f(n)}{n^s} \sum_{n\in\mathbb{N}} \frac{g(n)}{n^s}$ .

Here : $f(n)=\phi(n)$ & $g(n)=I(n)$

$\displaystyle\sum_{n\in\mathbb{N}} \underbrace{\frac{\underbrace{(\phi*I)(n)}_{\text{n}}}{n^s}}_{\zeta(s-1)} = \sum_{n\in\mathbb{N}} \frac{\phi(n)}{n^s} \sum_{n\in\mathbb{N}} \underbrace{\frac{\underbrace{I(n)}_{\text{1}}}{n^s}}_{\zeta(s)}$

$\displaystyle\implies \sum_{n\in\mathbb{N}} \frac{\phi(n)}{n^s} = \frac{\zeta(s-1)}{\zeta(s)}$

This gives the sum $\displaystyle \sum_{n=1}^\infty \frac{\phi(n)}{n^3}=\frac{\zeta(2)}{\zeta(3)}=\frac{\pi^2}{6\zeta(3)}$ making the answer $\boxed{36}$ .