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### Does n power of e grow much more faster than its Maclaurin polynomial? [duplicate]

I wonder how to calculate the following limit: $$\lim_{n\rightarrow\infty}\frac{1+n+\frac{{}n^{2}}{2!}+\cdots +\frac{n^{n}}{n!}}{e^{n}}$$ In the first sight, I think it should be zero, because ...
13k views

### Surprising identities / equations

What are some surprising equations / identities that you have seen, which you would not have expected? This could be complex numbers, trigonometric identities, combinatorial results, algebraic ...
81 views

### Summation of exponential series [duplicate]

Evaluate the limit: $$\lim_{n \to \infty}e^{-n}\sum_{k = 0}^n \frac{n^k}{k!}$$ It is not as easy as it seems and the answer is definitely not 1. Please help in solving it.
115 views

### Is the sequences$\{S_n\}$ convergent? [duplicate]

Let $$S_n=e^{-n}\sum_{k=0}^n\frac{n^k}{k!}$$ Is the sequences$\{S_n\}$ convergent? The following is my answer,but this is not correct. please give some hints. For all $x\in\mathbb{R}$, ...
129 views

### Weird limit $\lim \limits_{n\mathop\to\infty}\frac{1}{e^n}\sum \limits_{k\mathop=0}^n\frac{n^k}{k!}$ [duplicate]

$$\lim \limits_{n\mathop\to\infty}\frac{1}{e^n}\sum \limits_{k\mathop=0}^n\frac{n^k}{k!}$$ I thought this limit was obviously $1$ at first but approximations on Mathematica tells me it's $1/2$. Why ...
68 views

### Limits Problem : $\lim_{n \to \infty}[(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})]^{\frac{1}{n}}$ is equal to.. [duplicate]

Problem: How to find the following limit : $$\lim_{n \to \infty}[(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})]^{\frac{1}{n}}$$ is equal to (a) $\frac{4}{e}$ (b) $\frac{3}{e}$ (c) ...
284 views

### maybe this sum have approximation $\sum_{k=0}^{n}\binom{n}{k}^3\approx\frac{2}{\pi\sqrt{3}n}\cdot 8^n,n\to\infty$

prove or disprove this $$\sum_{k=0}^{n}\binom{n}{k}^3\approx\dfrac{2}{\pi\sqrt{3}n}\cdot 8^n,n\to\infty?$$ this problem is from when Find this limit ...
I'm trying to compute the following limit of a series: $$\lim_{n\to\infty} \sum_{k = n+1}^{\infty}\frac{e^{-n}n^{k}}{k!}$$ Factoring $e^{-n}$ out of the sum, applying the definition of $e^{x}$ as a ...
### $\lim_{n\to\infty} e^{-n}\sum_{k=1}^n \frac{n^k}{k!}$ [duplicate]
How can be evaluated this limit: $$\lim_{n\to\infty} e^{-n}\sum_{k=1}^n \frac{n^k}{k!} .$$ Thank you.