Question

20%20%5Cdisplaystyle%20%5Crm%20%5Csum_%7B%20%20n_%7B2%7D%20%3D%201%7D%5E%20%5Cinfty%20%20%20%5Cdots%20%5Cdisplaystyle%20%5Crm%20%5Csum_%7B%20%20n_%7B2022%7D%20%3D%201%7D%5E%20%5Cinfty%20%20%20%5Cfrac%7B1%7D%7Bn_%7B1%7D%20n_2%20%5Cdots%20n_%7B2022%7D%28n_%7B1%7D%20%2B%20n_2%20%20%2B%20%5Cdots%20%2B%20%20n_%7B2022%7D%29%7D%20" id="TexFormula1" title=" \displaystyle \rm \sum_{ n_{ 1} = 1}^ \infty \displaystyle \rm \sum_{ n_{2} = 1}^ \infty \dots \displaystyle \rm \sum_{ n_{2022} = 1}^ \infty \frac{1}{n_{1} n_2 \dots n_{2022}(n_{1} + n_2 + \dots + n_{2022})} " alt=" \displaystyle \rm \sum_{ n_{ 1} = 1}^ \infty \displaystyle \rm \sum_{ n_{2} = 1}^ \infty \dots \displaystyle \rm \sum_{ n_{2022} = 1}^ \infty \frac{1}{n_{1} n_2 \dots n_{2022}(n_{1} + n_2 + \dots + n_{2022})} " align="absmiddle" class="latex-formula">​

Answer 1

As a simpler example, consider the iterated sum with only 2 indices,

$\displaystyle \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \frac1{n_1n_2(n_1+n_2)}$

(The case with just one index is pretty simple, as it reduces to ζ(2) = π²/6.)

Let

$\displaystyle f(x) = \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \frac{x^{n_1+n_2}}{n_1n_2(n_1+n_2)}$

Differentiating and multiplying by x, we get

$\displaystyle x f'(x) = \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \frac{x^{n_1+n_2}}{n_1n_2} \\\\ = \left(\sum_{n_1=1}^\infty\frac{x^{n_1}}{n_1}\right) \left(\sum_{n_2=1}^\infty \frac{x^{n_2}}{n_2}\right) \\\\ = (-\ln(1-x))^2 = \ln^2(1-x)$

$\implies f'(x) = \dfrac{\ln^2(1-x)}x$

By the fundamental theorem of calculus (observing that letting x = 0 in the sum makes it vanish), we have

$f(x) = \displaystyle \int_0^x \frac{\ln^2(1-t)}t \, dt$

If we let x approach 1 from below, f(x) will converge to the double sum and

$\displaystyle \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \frac1{n_1n_2(n_1+n_2)} = \int_0^1 \frac{\ln^2(1-x)}x \, dx$

In the integral, substitute $x\mapsto1-x$, use the power series expansion for 1/(1 - x), and integrate by parts twice.

$\displaystyle \int_0^1 \frac{\ln^2(1-x)}x \, dx = \int_0^1 \frac{\ln^2(x)}{1-x} \, dx \\\\ = \sum_{m=0}^\infty \int_0^1 x^m \ln^2(x) \, dx \\\\ = \sum_{m=0}^\infty -\frac2{m+1} \int_0^1 x^m \ln(x) \, dx \\\\ = \sum_{m=0}^\infty \frac2{(m+1)^2} \int_0^1 x^m \, dx \\\\ = 2 \sum_{m=0}^\infty \frac1{(m+1)^3} \\\\ = 2 \sum_{m=1}^\infty \frac1{m^3} = 2\zeta(3)$

We can generalize this method to k indices to show that

$\displaystyle \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \cdots \sum_{n_k=1}^\infty \frac1{n_1n_2\cdots n_k(n_1+n_2+\cdots+n_k)} = (-1)^k \int_0^1 \frac{\ln^k(1-x)}x \, dx \\\\ = k!\,\zeta(k+1) = \Gamma(k+1)\zeta(k+1)$

Then the sum we want is

$\displaystyle \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \cdots \sum_{n_{2022}=1}^\infty \frac1{n_1n_2\cdots n_{2022}(n_1+n_2+\cdots+n_{2022})} = \boxed{\Gamma(2023)\zeta(2023)}$