Htykuut

Just for fun, I defined the following series. For $minBbb N$,
$$g_m=sum_{ngeq1}prod_{i=0}^{m}frac1{n+i}$$
I am seeking a closed form for $g_m$. I found $g_1, g_2$, and a general expression involving a lot of integrals.

$$
begin{align}
g_1=&sum_{ngeq1}frac1{n(n+1)}\
=&sum_{ngeq1}frac1nint_0^1 t_0^nmathrm{d}t_0\
=&sum_{ngeq1}int_0^1int_0^{t_0} t_1^{n-1}mathrm{d}t_1mathrm{d}t_0 \
=&int_0^1int_0^{t_0}sum_{ngeq0}t_1^{n}mathrm{d}t_1mathrm{d}t_0 \
=&int_0^1int_0^{t_0}frac{mathrm{d}t_1}{1-t_1}mathrm{d}t_0 \
=&1 \
end{align}
$$
Through the same process, it can be shown that $g_2=frac14$. In fact, through the same process, it can be shown that
$$g_m=int_0^1int_0^{t_0}int_0^{t_1}cdotsint_0^{t_{m-2}}frac{mathrm{d}t_{m-1}}{1-t_{m-1}}mathrm{d}t_{m-2}cdotsmathrm{d}t_2mathrm{d}t_1mathrm{d}t_0$$
Which I don't even know how to approach. Any ideas? Thanks.

This isn't a duplicate question, because I would like to know how to evaluate those nested integrals, if possible.

The sum telescopes:
$$frac{m}{x(x+1)cdots(x+m)}=frac1{x(x+1)cdots(x+m-1)}
-frac1{(x+1)(x+2)cdots(x+m)}.$$
Summing this from $1$ to $infty$ gives
$$msum_{n=1}^inftyfrac{1}{n(n+1)cdots(n+m)}
=frac1{(m-1)!}$$
so your sum is $1/m!$.

$sum_{n=1}^{infty}frac{1}{n(n+1)ldots(n+m)}=frac{1}{m}sum_{n=1}^{infty}(frac{1}{n(n+1)ldots(n+m-1)}-frac{1}{(n+1)(n+2)ldots(n+m)})$, which can be easily verified by adding the fractions. From here, you can define $V_{n}=frac{1}{n(n+1)ldots(n+m-1)}$, and so clearly

$$g_m=frac{1}{m}sum_{n=1}^{infty}(V_n-V_{n+1})$$ and from here it is fairly obvious what to do