Summation with negative factorial terms My problem arises in trying to prove that
$$\sum_{n=0}^{\infty}\frac{n\lambda^n}{n!}=\lambda e^\lambda$$
So what I did was
$$\sum_{n=0}^{\infty}\frac{n\lambda^n}{n!}=\sum_{n=0}^{\infty}\frac{\lambda^n}{(n-1)!}=\lambda\sum_{n=0}^{\infty}\frac{\lambda^{n-1}}{(n-1)!}=\lambda e^\lambda$$
, where I considered the taylos expansion of the exponential $e^x=\sum_{n=0}^{\infty}x^n / n!$ for the last step.
But I'm not okay with the logic here (even if my probable wrongdoings took me to the correct answer).
My main concern is that in the last sum we have a negative factorial in the denominator for the $n= 0$ term, and as far as I know this is an undefined quantity. We could avoid mentioning this by skipping this undefined term and doing $\sum_{n=1}^{\infty}$ instead of $\sum_{n=0}^{\infty}$ but how could then I justify me ignoring a term that is possibly not zero in that series?
I guess my question can be rephrased as why the Taylor expansion of $e^x$ can be both $\sum_{n=0}^{\infty}x^n / n!$ and $\sum_{n=0}^{\infty}x^{(n-1)} / (n-1)! = \sum_{k=-1}^{\infty}x^k / k!$ at the same time?
 A: The first term of the series is $$\frac{0\times\lambda^0}{0!}=0$$
The rest of the series is $$\lambda\sum_{n=1}^\infty\frac{\lambda^{n-1}}{(n-1)!}$$
Now re-index this with $i=n-1$ to give $$\lambda\sum_{i=0}^\infty\frac{\lambda^i}{i!}=\lambda e^{\lambda}$$
A: Note that $\Gamma(n)$, for $n \leq 0$ is essentially $\pm \infty$ and gives $\dfrac{1}{\Gamma(-n)} = 0$. This applies to the series as
\begin{align}
\sum_{n=0}^{\infty} \frac{n \, t^n}{n!} &= \sum_{n=0}^{\infty} \frac{t^n}{(n-1)!} \\
&= \frac{1}{(-1)!} + \sum_{n=1}^{\infty} \frac{t^n}{(n-1)!} \\
&= \sum_{n=1}^{\infty} \frac{t^n}{(n-1)!} = t + \frac{t^2}{1!} + \cdots \\
&= t \, \left( 1 + \frac{t}{1!} + \frac{t^2}{2!} + \cdots \right) \\
&= t \, \sum_{n=0}^{\infty} \frac{t^n}{n!} = t \, e^t.
\end{align}
In a similar manor consider the series:
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2 \, t^n}{n!} &= \sum_{n=0}^{\infty} \frac{(n(n-1) + n) \, t^n}{n!} \\
&= \sum_{n=0}^{\infty} \frac{t^n}{(n-2)!} + \sum_{n=0}^{\infty} \frac{t^n}{(n-1)!} \\
&= \frac{1}{(-2)!} + \frac{t}{(-1)!} + \sum_{n=2}^{\infty} \frac{t^n}{(n-2)!} + t \, e^t \\
&= \sum_{n=0}^{\infty} \frac{t^{n+2}}{n!} + t \, e^t \\
&= t \, (t+1) \, e^t.
\end{align}
Alternatively, using differential operators,
\begin{align}
\sum_{n=0}^{\infty} \frac{n \, t^n}{n!} &= t \, \sum_{n=0}^{\infty} \frac{n \, t^{n-1}}{n!} \\
&= t \, \frac{d}{dt} \, \sum_{n=0}^{\infty} \frac{t^n}{n!} = t \, \frac{d}{dt} \left( e^t \right) \\
&= t \, e^t.
\end{align}
