These notes on multiplicative number theory mention the convolution $ 1 \ast 1 = \tau$ (where $\tau$ is the divisor function not Ramanujan tau function. Therefore
$$ \bigg(\sum \frac{1}{n^s} \bigg)^2 = \sum \frac{\tau(n)}{n^s}$$
At this point, I would like to plug in the value $s =0$ we should get that:
$$ \sum \tau(n) = \bigg(1+1+1+\dots \bigg)^2 = \bigg( -\frac{1}{2} \bigg)^2 = \frac{1}{4}$$
We could also plug in $s=-1$ to get another relation betweeen divergent sums:
$$ \sum n \tau(n) = \bigg(1+2+3+\dots \bigg)^2 = \bigg( -\frac{1}{12} \bigg)^2 = \frac{1}{144}$$
One Wikipedia, there is some discussion of the divergent series $1+1+1+\dots$ but the result involves some very complicated analytic continuation.
Are there any simpler ways or summation methods to handle these cases?
How do we compute the square of $\zeta(s)$ ?
$$ \zeta(s)^2 = \bigg(\sum \frac{1}{a^s} \bigg) \bigg(\sum \frac{1}{b^s} \bigg) = \sum_{n =1}^\infty \sum_{a \times b=n} \frac{1}{(ab)^s} = \sum \frac{\tau(d)}{n^s} $$
How many copies of $\frac{1}{n^s}$ ? One for each factorization $n = a \times b$, this number is called $\tau(d)$.
Example: $18 = 1 \times 18 = 2 \times 9 = 3 \times 6 = 6 \times 3 = 9 \times 2 = 18 \times 1 $. So there should be 6 factors, $\frac{6}{18^s}$.
There is also must formula for multiplying Dirichlet series. Sorry I am running out of letters.
$$ \bigg(\sum \frac{a(n)}{n^s} \bigg) \bigg(\sum \frac{b(m)}{m^s} \bigg) = \sum \bigg( \sum_{m \times n = r} a(m)b(n)\bigg)\frac{1}{r^s} $$