Please help me out to find the limit of this series $$\lim_{n\to \infty} \sum_{i=0}^n e^{-n} \frac{n^i}{i!} $$  Upper Limit is easy to find , how to find exact limit
5228
I'm supposed to calculate: $$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}$$ By using WolframAlpha, I might guess that the limit is $\frac{1}{2}$, which is a pretty interesting and nice result. I wonder in which ways we may approach it.
2235
Can someone show me how to calculate the limit:  $$\lim_{x \to \infty }\left(\frac{1}{e}\left(1+\frac{1}{x}\right)^x\right)^x $$  I tried to use taylor series but failed.  Thanks
114120
How can I calculate this: $\lim\limits_{n \rightarrow \infty} \frac{1+\sqrt[n] {e}+\sqrt[n] {e^2}+ \dots + \sqrt[n] {e^{n-1}}  } {n}$? I how come so far: $$ \lim_{n \rightarrow \infty} \frac{1+\sqrt[n] {e}+\sqrt[n] {e^2}+ \dots + \sqrt[n] {e^{n-1}}  } {n} = \lim_{n \rightarrow \infty} \frac{1}{n} \left(1 +\sqrt[n] {e}+\sqrt[n] {e^2}+ \dots + \sqrt[n] {e^{n-1}}\right) = $$  $$ \lim_{n \rightarrow \infty}\frac{1}{n}\sum_{i =1}^{n}e^{\frac{i-1}{n}}$$ But how do I continue?
16311
As seen in  question, the series $\sum\limits_{n=1}^{\infty} \dfrac{n!e^n}{n^n}$ diverges. (One way to see this is by noting that the terms of the sum are greater than $1$ and, therefore, don't converge to zero.)  However, more is true: not only are the terms greater than $1$, they blow up to $+\infty$! To see this, one can apply Stirling's approximation and get $\dfrac{n!e^n}{n^n} \sim \sqrt{2\pi n}$.  My question is: is there an elementary proof for $\lim_{n \to\infty}\dfrac{n!e^n}{n^n} = +\infty$?  The reason I'm asking this is that I'm taking a Calculus course and I was assigned the exercise of proving the divergence of the series above and I would feel more satisfied if I could prove that its terms go to $+\infty$ (as opposed to merely proving the terms are greater than $1$.) Obviously, things like Stirling's approximation are outside the scope of the course, so I can't really use them.  (To be clear, you can use anything one learns in $4$ semesters of standard Calculus courses. I hope this restriction is not too obscure.)
56463
How can we solve the following?:  $$\lim_{N\to \infty}  \sum_{r=1}^N \frac{2^r}{(1+5^{2^r})} $$  The answer according to Wolfram Alpha is $\frac{1}{12}$, but I have no idea how it can be calculated on paper without summing it infinitely (which apparently is what Wolfram Alpha does).
100991
I have tried to show that this limit : $$\lim\limits_{n\to \infty }\frac{n}{n!^{\frac 1 n}}=e$$  using  $ \lim  (1+\frac 1 n)^{\frac 1 n} , n \to \infty $ , I don't find any equivalence , however wolfram alpha says that is $e$ as shown , then how do I evaluate it ?
93825
$\lim \limits_{x \to 1^-} \displaystyle \sum_{n=0}^\infty (-1)^nx^{n²} = \frac{1}{2}$  The power $n^2$ is problematic. Can we bring this back to the study of usual power series?  I do not really have any idea for the moment.
5928
$$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k+n}$$  Could you please give me a hint how to find this limit?
108692
This is the series $\sum_{k=1}^{\infty} \frac{7^{k} + k}{k! + 1}$. I began with the ratio test, but after I set up the ratio to take the limit of, I couldn't simplify the limit, since nothing would cancel out. I believe it converges, but I have no idea how to solve further after setting up that ratio test limit.  If someone has another method, that would be fine too. We are not allowed to use Comparison or Limit Comparison Tests.
115761
I am trying to find this limit:  $$\lim_{n\to\infty}\left(n^{\tfrac{1}{n}}-1\right)^n,$$  I tried using exponential function, but I see no way at the moment. I am not allowed to use any kind of differentiation or other topics of advanced math, only induction and school math are possible. Thank you
57789
Inspired by the exponential series, I'm curious about where exactly the series $\displaystyle\sum_{n=1}^\infty\frac{z^n}{n}$ for $z\in\mathbb{C}$ converges.   I calculated $$ \limsup_{n\to\infty}\sqrt[n]{\frac{1}{n}}=\limsup_{n\to\infty}\frac{1}{n^{1/n}} $$ and $$ \lim_{n\to\infty} n^{1/n}=e^{\lim_{n\to\infty}\log(n)/n}=e^{\lim_{n\to\infty}1/n}=e^0=1. $$ So the radius of convergence is $1$, so the series converges on all $z$ inside $S^1$. But is there a way to tell for which $z$ on the unit circle the series converges? I know it converges for $z=-1$, but diverges for $z=1$, but I don't know about the rest of the circle. For what other $z$ does this series converge? Thanks.
31748
How to find the value of $\lim_{n\to\infty}S(n)$, where $S(n)$ is given by $$S(n)=\displaystyle\sum_{k=1}^{n} \dfrac{k}{n^2+k^2}$$   is unable to calculate it.  This is a question from a questions booklet, and the options for the answer are--  $\begin{align} &amp;A) \dfrac{\pi}{2} \\ &amp;B) \log 2 \\ &amp;C) \dfrac{\pi}{4} \\ &amp;D) \dfrac{1}{2} \log 2 \end{align}$
20256
How can I compute the following limit? $$ \lim_{n\to\infty}\frac{\dfrac{n}{1}+\dfrac{n-1}{2}+\dots+\dfrac{2}{n-1}+\dfrac{1}{n}}{\ln(n!)} $$  I have tried lots of methods, I can't get the answer. Although I think the limit is $0$, I don't know how to explain it. Please, if someone could help me it would be fantastic.
113177
assume  $$x_n=\frac{n+1}{2^{n+1}}\sum_{k=1}^n\frac{2^k}{k} ,n=1,2,.....$$  how compute $\lim_{n\to +\infty}x_n$?  Thanks for any hint
3701
I was wondering if the following is true. Suppose a series $a_n$ is greater than 0 for all positive integer n, and that $\sum \frac {a_n}n$ converges, then is $\displaystyle \lim_{m\to \infty}\sum_{n= 1}^m {a_n \over m+n} = 0$? It seems to be true because if $\sum {a_n\over n}$ converges, then that means that ${a_n\over n }\to 0$ for $n\to \infty$. This means, neglecting $n$, ${a_n \over m+n}$ will also tend to 0, and thus the summation would be equal to 0, but I don't know if this is true.
89503
Does the series $\sum \limits_{n=1}^{\infty} \sin(n^2)$ converge?  I think I've tried everything, I have no more ideas.
68775
Who knows about the summation of this series: $$\sum\limits_{i=1}^{n}i^m $$  where $m$ is constant and $m\in \mathbb{N}$?  thanks