Strange equality involving a geometric series and gamma and zeta function

Question

I saw someone do this (in a youtube video):

$$\sum_{\text{n}=1}^\infty\frac{\Gamma\left(\text{s}\right)}{\text{n}^\text{s}}=\Gamma\left(\text{s}\right)\sum_{\text{n}=1}^\infty\frac{1}{\text{n}^\text{s}}=\Gamma\left(\text{s}\right)\zeta\left(\text{s}\right)=\sum_{\text{n}=1}^\infty\left\{\int_0^\infty\text{u}^{\text{s}-1}e^{-\text{n}\text{u}}\space\text{d}\text{u}\right\}=$$ $$\int_0^\infty\text{u}^{\text{s}-1}\left\{\sum_{\text{n}=1}^\infty e^{-\text{n}\text{u}}\right\}\space\text{d}\text{u}=\int_0^\infty\text{u}^{\text{s}-1}\cdot\frac{1}{e^\text{u}-1}\space\text{d}\text{u}$$

But, I can follow all the steps he did but the last integral does not converge because the geometric series only hold when the real part of $\text{u}$ is bigger then $0$, but the lower bound of the integral equals $0$. So why are those two things equal?

Or can we assign a value to:

$$\lim_{u\to0}\text{u}^{\text{s}-1}\cdot\frac{1}{e^\text{u}-1}$$

Answer 1

Last statement could wrote like this ‎$$\frac{u^{s-1}}{e^u-1}=\frac{e^{-u}u^{s-1}}{1-e^{-u}}=e^{-u}u^{s-1}\sum_{k=0}^{\infty}e^{-ku}=\sum_{k=1}^{\infty}e^{-ku}u^{s-1}$$‎

Answer 2

Usually, we take integrals like this as follows:

$$\int_0^\infty\frac{u^{s-1}}{e^u-1}\ du=\lim_{a\to0}\int_a^1\frac{u^{s-1}}{e^u-1}\ du+\lim_{b\to\infty}\int_1^b\frac{u^{s-1}}{e^u-1}\ du$$

And when $s>1$, there is no problem as $a\to0$.