Sums of harmonic series If $$f(n, a, b)=\sum_{k_1=a+1}^{b}\sum_{k_2=a+1}^{k_1}\cdots\sum_{k_{2n}=a+1}^{k_{2n-1}}\frac{1}{k_1k_2\ldots k_{2n}}$$ $f(0, a, b)=1$ and $$g(a, b)=\sum_{n=0}^{\infty}f(n, a, b)$$ is there a way to find an exact value for $g(a, b)$
I had thought of changing the sums to integrals to get an approximation, but I was thinking if there is an explicit expression.
 A: Not a full answer
First I'll change the function to an easier to deal with function:
$$q(n,a,b) = \sum_{k_1=a}^{b}\sum_{k_2=a}^{k_1}\cdots\sum_{k_{n}=a}^{k_{n-1}}\frac{1}{k_1k_2\ldots k_{n}}$$
$$f(n,a,b) = q(2n,a+1,b)$$
$$q(n,a,b) = \sum_{k_1=a}^{b}\frac{1}{k_1}\sum_{k_2=a}^{k_1}\frac{1}{k_2}\cdots\sum_{k_{n}=a}^{k_{n-1}}\frac{1}{k_{n}}$$
Now, I'll use the generalized harmonic numbers:
$$H_{n,m}= \sum^n_{k=0}\frac{1}{k^m}$$
let's define $q$ using a recurrence relation:
$$q(n,a,b) = \sum^b_{k=a}\frac{1}{k}g(n-1,a,k)$$
Some small examples
$$q(0,a,b)=1$$
$$q(1,a,b)=\sum^b_{k=a}\frac{1}{k}g(0,a,k)=\sum^b_{k=a}\frac{1}{k}=H_{b,1}-H_{a,1}$$
$$q(2,a,b)=\sum^b_{k=a}\frac{1}{k}g(1,a,k)=\sum^b_{k=a}(H_{k,1}-H_{a,1})\frac{1}{k}=\sum^b_{k=a}\frac{H_{k,1}}{k}-({(H_{a,1})}^2-H_{a,1}H_{b,1})$$
from this:Wolfram
$$\frac{1}{2}({(H_{b,1})}^2+H_{b,2}-{(H_{a,1})}^2-H_{a,2})-({(H_{a,1})}^2-H_{a,1}H_{b,1})=\frac{{(H_{b,1})}^2+H_{b,2}-H_{a,2}}{2}+\frac{3}{2}{(H_{a,1})}^2 + H_{a,1}H_{b,1}$$
$$$$
from here, it seems that $g(3,a,b)$ doesn't have an expression of this form, as the sum of $\frac{H_{2,k}}{k}$ doesn't seem to be expressed as such.
