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The displayed definition uses iterated Eisenstein summation in weight two: the sum in $n$ is evaluated before the sum in $m$, with the single $(0,0)$ term omitted. This order is essential. The two-dimensional lattice series does not have absolute convergence, so arbitrary rearrangement would not be justified.

For noninteger $w$, the cosecant partial-fraction identity isFor completeness, apply the residue theorem to $\pi \cot (\pi \zeta )/(\zeta -w{)}^2$ on squares with large half-integer sides. The cotangent is bounded on the contours and the integral is $O(R^{-1})$. Its residues at the integers are $(n-w{)}^{-2}$ and its residue at $w$ is $-{\pi }^2{\csc }^2(\pi w)$, proving the formula. If $\mathrm{Im}w>0$, the geometric-series expression $\cot (\pi w)=-i(1+2{\sum }_{r\ge 1}{e}^{2\pi irw})$, differentiated termwise, gives the cotangent partial-fraction Fourier kernel

Put $q=e^{2\pi iz}$ with $z$ in the complex upper half-plane. For positive $m$, this gives $-4{\pi }^2{\sum }_{r\ge 1}r{q}^{mr}$. Negative $m$ gives the same value, by replacing $n$ with $-n$ in its inner sum. The $m=0$ row is ${\sum }_{n\ne 0}{n}^{-2}={\pi }^2/3$, by the Basel problem. The resulting series in $m,r$ does have absolute convergence, locally uniformly in $z$, so collecting the coefficient at $q^{\ell }$ is legitimate:The coefficient is the sum-of-divisors function, since $r$ runs over the positive divisors of $\ell$. ThusThere is no conflict with vanishing of weight-two level-one modular forms: the Eisenstein series of weight two has an anomalous transformation term, so it is not a weight-two modular form.

An arithmetic function $f$ is multiplicative if $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are coprime integers. If $g(n)={\sum }_{d\mid n}f(d)$ and $(m,n)=1$, every positive integer divisor of $mn$ is uniquely $d_1{d}_2$ with $d_1\mid m$ and $d_2\mid n$. Henceso $g$ is multiplicative.

The Möbius function is defined by $\mu (1)=1$, $\mu (n)=0$ if the square of a prime number divides $n$, and $\mu (n)=(-1{)}^r$ if $n$ is a product of $r$ distinct prime numbers. Coprime integers have disjoint sets of prime factors, so this definition immediately gives $\mu (mn)=\mu (m)\mu (n)$ when $(m,n)=1$.

For $n=p_1^{a_1}\cdots p_r^{a_r}>1$, only the square-free divisors contribute, and thereforeFor $n=1$ the sum is $1$. Thus $1\ast \mu =\epsilon$ in the language of Dirichlet convolution. Convolving $g=1\ast f$ with $\mu$ gives the Möbius inversion formula

If $f(n)=n$, its divisor sum is the sum-of-divisors function. Conversely, if $g(n)=n$, Möbius inversion givesusing the standard divisor identity for the Euler totient function. Therefore the two requested answers are