Suppose we seek to evaluate
$$S(N) = \sum_{n=N}^\infty
\sum_{k=N}^n \sum_{j=0}^k {k\choose j} (-1)^{k-j}
(1+j)^n \frac{t^n}{n!}.$$
This is
$$S(N) = \sum_{n=N}^\infty \frac{t^n}{n!}
\sum_{k=N}^n \sum_{j=0}^k {k\choose j} (-1)^{k-j}
(1+j)^n.$$
Now introduce
$$(1+j)^n =
\frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} \exp((1+j)z) \; dz.$$
We thus get for the inner sum
$$\frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\sum_{j=0}^k {k\choose j} (-1)^{k-j}
\exp((1+j)z) \; dz
\\ = \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{n+1}}
(\exp(z)-1)^k\; dz.$$
Substitute this into the middle sum to get
$$\frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{n+1}}
\frac{(\exp(z)-1)^{n+1} - (\exp(z)-1)^N}{\exp(z)-2}
\; dz.$$
Now since $\exp(z)-1$ starts at $z$ the first term drops out and we
get
$$\frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{n+1}}
\frac{(\exp(z)-1)^N}{2-\exp(z)}
\; dz.$$
We thus get for the remaining sum
$$\sum_{n=N}^\infty \frac{t^n}{n!} \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{n+1}}
\frac{(\exp(z)-1)^N}{2-\exp(z)}
\; dz.$$
Finally note that $(\exp(z)-1)^N$ starts at $z^N$ so we may lower
the initial value of the remaining summation to zero, getting
$$\sum_{n=0}^\infty t^n \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{n+1}}
\frac{(\exp(z)-1)^N}{2-\exp(z)}
\; dz.$$
What we have here is an annihilated coefficient extractor which
finally yields
$$\frac{\exp(t)}{2-\exp(t)} (\exp(t)-1)^N.$$
Now for the exponential growth rate of the coefficients on $t^n$ we
get the distance to the nearest singularity which is $\log 2.$ So the
radius of convergence of this sum is $|t|<\log 2.$
Observation. Having reached the end of this computation we see that we didn't need to substitute the variable in the integral, which means we could have used the coefficient extractor notation $[z^n]$ throughout. This does not affect the semantics of the computation.
Remark. There are several more examples of the technique of
annihilated coefficient extractors at this MSE link
I and at this MSE
link II and also
here at this MSE link
III.