I realise this is quite simple, but I just don't exactly understand what the last step is. $$\sum_{n=1}^\infty\frac{n}{(n-1)!}=\sum_{n=1}^\infty\left(\frac{1}{(n-2)!}+\frac{1}{(n-1)!}\right)=\sum_{n=1}^\infty\frac{1}{(n-2)!}+e$$
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In the last step, what happened is that the sum was broken into two. If you can prove the series converges, this is justified:
$$\sum_{n=1}^\infty\left(\frac{1}{(n-2)!}+\frac{1}{(n-1)!}\right) = \sum_{n=1}^\infty \frac{1}{(n-2)!}+ \sum_{n=1}^\infty \frac{1}{(n-1)!} $$
Compare this last series with the Taylor series for $e^x$:
$$e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$$
PrincessEev
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But how can I evaluate the sum $\sum_{n=1}^\infty\frac{1}{(n-2)!}$ if (-1)! cannot be evaluated? – Jakub Gluszko Feb 27 '23 at 20:49
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1Think of $\dfrac{1}{(n-2)!}$ as $\dfrac{(n - 1)}{(n - 1)!}$ (which is where it comes from in the first step); the term is zero when $n = 1$, and so rewriting it as $\dfrac{1}{(n-2)!}$ is really only meaningful is $n \neq 1$. – Jakob Streipel Feb 27 '23 at 20:56
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Ah that makes sense. Thanks. – Jakub Gluszko Feb 27 '23 at 20:57