An infinite summation of Totient.

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

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

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

$\displaystyle\sum_{n=1}^{\infty} \frac{\phi(n)}{n^4} = \frac{\zeta{(4-1)}}{\zeta{(4)}}=\frac{90}{\pi^4}\zeta{(3)}$

$\boxed{a+b+c=97}$

Let $F(s) = \displaystyle \sum_{n=1}^{\infty} \dfrac{\phi(n)}{n^s}$

Using $\phi(n) = \displaystyle \sum_{d|n} \mu(d) \dfrac{n}{d}$ , and reversing the order of summation , we have:

$F(s) = \displaystyle \sum_{d=1}^{\infty} \dfrac{\mu(d)}{d} \sum_{n:d|n}\dfrac{1}{n^{s-1}}$

Substituting $n = m d$ ,

$F(s) = \displaystyle \sum_{d=1}^{\infty} \dfrac{\mu(d)}{d^s} \zeta(s) = \boxed{\dfrac{\zeta(s-1)}{\zeta(s)}}$

$S=\prod_{p}\sum_{k=0}^{\infty}\frac{\phi(p^k)}{p^{4k}}=\prod_{p}\left(1+\sum_{k=1}^{\infty}\frac{p-1}{p^{3k+1}}\right)=\prod_{p}\frac{p^4-1}{p^4-p}=\prod_{p}\frac{1-\frac{1}{p^4}}{1-\frac{1}{p^3}}=\frac{\zeta(3)}{\zeta(4)}=\frac{90\zeta(3)}{\pi^4}$ giving $\boxed{97}$