Linked Questions

2
votes
1answer
45 views

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 ...
2
votes
1answer
92 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.
1
vote
1answer
67 views

Asymptotics of $\sum_{i=0}^{n} \frac{i n^i}{i!}$.

How can you calculate the asymptotics of $\sum_{i=0}^{n} \frac{i n^i}{i!}$ ? This sum looks similar to the power series for $e^n$. I have also seen similar problems solved using Poisson distributions ...
6
votes
0answers
291 views

Finding $\lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}$ if it exists [duplicate]

Does there exist the following limitation? If the answer is yes, could you show me how to find that? $$\lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}$$ In the following, I'm going to write what ...
2
votes
0answers
61 views

$ \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.
1
vote
0answers
55 views

Limit Challenge [duplicate]

I have tried very hard to manipulate this limit, but can't seem to find what the answer is. Can someone give me an explanation to what the limit would be. $$\lim_{n\rightarrow \infty} e^{-n}\cdot ...
1
vote
0answers
33 views

Compare $e^n$ and its first $n$ terms sum [duplicate]

Compute the limit as $n$ approaches infinity. $$ \frac{\sum_{0\le i\le n} \frac{n^i}{i!}}{e^n} $$ It is somehow between 0 and 1.
0
votes
0answers
29 views

Limit of Series with Variable Lower Bound [duplicate]

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 ...

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