Stieltjes constants are defined as:
$$\gamma_n=\lim_{m\to\infty}\left[ \left( \sum_{k=1}^m\frac{(\ln k)^n}{k}\right)-\frac{(\ln m)^{n+1}}{n+1} \right]$$
I want to compute the following series:
$$\begin{align} \sum_{n=0}^\infty \frac{\gamma_n}{n!}&=\lim_{m\to\infty}\left[ \left( \sum_{k=1}^m\sum_{n=0}^\infty\frac{(\ln k)^n}{n!~k}\right)-\sum_{n=0}^\infty\frac{(\ln m)^{n+1}}{n!~(n+1)} \right]\\ \\ &=\lim_{m\to\infty}\left[ \left( \sum_{k=1}^m\frac{e^{\ln(k)}}{k}\right)-\sum_{n=0}^\infty\frac{(\ln m)^{n+1}}{(n+1)!} \right]\\ \\ &=\lim_{m\to\infty}\left[ m-\sum_{p=1}^\infty\frac{(\ln m)^{p}}{p!} \right]~~~~~~~~~~~(p=n+1)\\ \\ &=\lim_{m\to\infty}\left[ m-\left(e^{\ln(m)}-1 \right) \right]\\ \\ &=\lim_{m\to\infty}\left[ 1 \right]\\ \\ &=1 \end{align}$$
But the correct answer is $\frac{1}2$...
I wouldn’t say that this is “the” definition of the constants. Much more usefully (to this purpose) they are defined to be the Laurent series coefficients of $\zeta$: $$\zeta(s)=(s-1)^{-1}+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\gamma_n(s-1)^n$$Which, if we evaluate at zero, gives: $$\begin{align}-\frac{1}{2}&=(-1)^{-1}+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\gamma_n(-1)^n\\&=-1+\sum_{n=0}^\infty\frac{\gamma_n}{n!}\\\therefore\frac{1}{2}&=\sum_{n=0}^\infty\frac{\gamma_n}{n!}\end{align}$$As $\zeta(0)=-1/2$ is known.
As Brian notes, your error was in an unjustified limit interchange.