Is this result correct? I have the following property of summation $e^{x}=\sum_{k=0}^{∞}\frac{x^{k}}{k!}$, then $$\sum _{k=61} ^{\infty} \frac{e^{-66}\left(66\right)^{k}}{k!} = \frac{1}{e^{66}} \sum_{k=61} ^{\infty} \frac{\left(66\right)^{k}}{k!}=\frac{e^{66}}{e^{66}}=1$$
The correct answer is supposed to be $0.747$ but I have not found a structure or a recursive method to know where the result comes from.
Thank you very much for anyone who can give me a good indication.
 A: So, alluding to what some of the comments are saying, you made the mistake in your work assuming $$\sum_{k=61}^\infty \frac{(66)^k}{k!} = e^{66}$$
where $k$ starts at $k=61$. This is incorrect. You would have $e^{66}$ exactly if you started the series from $k=0$. That is, $e^{66} = \sum_{k=0}^\infty \frac{(66)^k}{k!}$.
Furthermore, this is not what the answer did in the question you linked in the above comments. In that question, they had $$\sum_{n=2}^{\infty} \frac{e^{-2} 2^{n}}{n!} = e^{-2} \left(e^{2} - 1 - 2 \right)$$
Although they did start at $n=2$, notice they also had to do some adjustments on the right-hand side of the equation. With that being said, to answer your question, we do want to still use the series expansion for the function $e^x$. You started out correctly,
$$\sum _{k=61} ^{\infty} \frac{e^{-66}\left(66\right)^{k}}{k!} = \frac{1}{e^{66}} \sum_{k=61} ^{\infty} \frac{\left(66\right)^{k}}{k!} \ \ \ \ (\text{equation }1)$$
is a good start. The question is then, what do we do with the factor $\sum_{k=61} ^{\infty} \frac{\left(66\right)^{k}}{k!}$ on the right-hand side of equation $1$? Well, observe that we may replace it with $e^{66} - \sum_{k=0}^{60} \frac{(66)^k}{k!}$. Thus, $$\sum _{k=61} ^{\infty} \frac{e^{-66}\left(66\right)^{k}}{k!} = \frac{1}{e^{66}} \sum_{k=61} ^{\infty} \frac{\left(66\right)^{k}}{k!} = \frac{1}{e^{66}}\left(e^{66} - \sum_{k=0}^{60} \frac{(66)^k}{k!} \right) = 1 - \frac{1}{e^{66}}\sum_{k=0}^{60} \frac{(66)^k}{k!}$$
Admittedly although this still looks rather nasty, we have already made a huge improvement because we've been able to switch out an infinite series for a finite sum. As long as you can compute the finite sum $\frac{1}{e^{66}}\sum_{k=0}^{60} \frac{(66)^k}{k!}$ (by hand if you have a lot of time haha or with a quick program on Python), you'll be good to go.
