I want to calculate      $$\lim_{n\to \infty} n\sin(2\pi n! e)$$   I have used the Stirling approximation and I think the answer is zero . But I think the limit maybe not exists.   Can some one help?  Thanks.
17841
I tried and got this   $$e=\sum_{k=0}^\infty\frac{1}{k!}=\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{k!}$$ $$n!\sum_{k=0}^n\frac{1}{k!}=\frac{n!}{0!}+\frac{n!}{1!}+\cdots+\frac{n!}{n!}=m$$ where $m$ is an integer. $$\lim_{n\to\infty}n\sin(2\pi en!)=\lim_{n\to\infty}n\sin\left(2\pi n!\sum_{k=0}^n\frac{1}{k!}\right)=\lim_{n\to\infty}n\sin(2\pi m)=\lim_{n\to\infty}n\cdot0=0$$  Is it correct?
49586
Assume  $$\lim_{n\to \infty}  (y_{2n} - y_n)=0$$  then  $$\lim_{n\to \infty} y_n$$  exists.  I know it's not true, and I can see a sequence that disprove that $(1,1,2,1,3,2,4,1,5,3,...)$ but I want a sequence that has formula.  hint/help?  thanks
65752
I need to find the limit for:   $ \lim_{n\to\infty} \frac {(n!)^\frac{1}{n}}{n} $  I know the answer is $\frac {1}{e}$ but I have no idea how to get that answer. I'd appreciate some help.
77797
Compute $$\displaystyle \lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \sum_{k=1}^n (\sin{x})^kdx.$$   I tried writing $\displaystyle \lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \sum_{k=1}^n (\sin{x})^kdx = \lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \frac{1-(\sin{x})^{n+1}}{1-\sin{x}}dx$, but I don't know how to continue form here.  I know that the limit diverges to $\infty$, but I don't know how to prove it.
101292
$$\lim_{x \to \frac{2}{\pi}}\lfloor \sin \frac{1}{x} \rfloor=0$$    I need to prove the limit of this using the $\epsilon - \delta $ way but I don't know how to find $\delta$ when I'm given a trigonometric function I know only how to do it with polynomial functions
95514
I'm trying to find some nice proofs for the following limit  $$\lim_{n\to\infty} \int_{0}^{\pi/4}\tan^n x \ dx$$  One way is to use the integration by parts. What else can we do here? Are there some fast ways? All answers will have my upvote. Thanks!
107633
I tried to calculate following limit: (let me present my solution first, and then point errors or give hints how to finish)$$\lim_{n\rightarrow\infty}\frac{(n!)^n}{n^{{n}^{2}}}$$So my partly solution is: Let $a_{n}=\frac{(n!)^n}{n^{{n}^{2}}}$. Then $a_{n}=(\frac{(n!)}{n^{n}})^n$ and denoting $p_{n}=ln (a_{n})$ we have $p_{n}=ln((\frac{(n!)}{n^{n}})^n)$$=n*ln\frac{(n!)}{n^{n}}=$$n*(ln(n!)-n*ln(n)$ and now I use Stirling formula  $n!\approx\sqrt{2\pi n}*(\frac{n}{e})^n$ and now $p_{n}=n*(ln(\sqrt{2\pi n}*(\frac{n}{e})^n)-n*ln(n))$$=n(ln(\sqrt{2\pi n)}+n*ln(n)-n-n*ln(n))=$$=n((ln(\sqrt{2\pi n})+n))=$$=n(ln(n\sqrt{{2n\pi n})})$ and that would imply that $\lim_{n\rightarrow\infty}a_{n}=exp(n\sqrt{2\pi n})^n$ what does not seem good... Where am I doing mistake, is it wrong way of acting? Any hints are desired, thank you in advance.
101295
I need to study this sum: $\sum_{n=1}^{\infty} \frac{n!\, i^n}{n^n}$. Taking: $$\lim_{n\rightarrow \infty} \; \left|\left(\frac{n!\: i^n}{n^n}\right)^{\frac{1}{2}}\right|$$ $$\rightarrow \lim_{n\rightarrow \infty}  \left|\left(\frac{n!}{n^n}\right)^{\frac{1}{2}}\right|$$  Using Stirling approximation considering the limit:  $$\ln n! \approx n \ln\:n-n$$ then $n! \approx (\frac{n}{e})^n$ (is this correct?):  $$\rightarrow \lim_{n\rightarrow \infty} \left|\frac{n}{n\, e}\right| = \frac{1}{e}$$ This doesn't make too much sense because I know the series diverges.  I think I'm missing a $\sqrt{2\pi n}$ in Stirling approximation but I don't understand why that pops out taking the $e^{(\;)}$ from the first expression.
63519
$$\lim_{n\to\infty} (1+\frac{2}{n})^{n^2} e^{-2n} = ? $$         $e$   $e^2$   $e^{-1}$   $e^{-2}$      My answer is 1. Since $$\lim_{n \to \infty} \left(1+\frac{2}{n}\right)^{n} =e^{+2}.$$  Therefore $$\lim_{n\to\infty} \left(1+\frac{2}{n}\right)^{n^2} =e^{2n}.$$  Where am I wrong?
8977
$$\lim_{n \to\infty} \frac{n!}{n^n}$$  I have a question: is it valid to use Stirling's Formula to prove convergence of the sequence?
33867
I was asked today by a friend to calculate a limit and I am having trouble with the question.  Denote $\sin_{1}:=\sin$ and for $n&gt;1$ define $\sin_{n}=\sin(\sin_{n-1})$. Calculate $\lim_{n\to\infty}\sqrt{n}\sin_{n}(x)$ for $x\in\mathbb{R}$ (the answer should be a function of $x$ ).  My thoughts:   It is sufficient to find the limit for $x\in[0,2\pi]$ , and it is easy to find the limit at $0,2\pi$ so we need to find the limit for $x\in(0,2\pi)$.  If $[a,b]\subset(0,\pi)$ or $[a,b]\subset(\pi,2\pi)$ we have it that then $$\max_{x\in[a,b]}|\sin'(x)|=\max_{x\in[a,b]}|\cos(x)|&lt;\lambda\leq1$$ hence the map $\sin(x)$ is a contracting map.  We know there is a unique fixed-point but since $0$ is such a point I deduce that for any $x\in(0,2\pi)$ s.t $x\neq\pi$ we have it that $$\lim_{n\to\infty}\sin_{n}(x)=0$$  So I have a limit of the form "$0\cdot\infty$" and I can't figure out any way on how to tackle it.  Can someone please suggest a way to find that limit ?  Note: I am unsure about the tags, please change them if you see fit.
43522
I think should be something very simple... $$\lim_{n\to\infty}nq^{n}$$ wher $|q|&lt;1$ i tried to use binomial theorem, but no success... I know that $q^{n}$ is convergent to $0$, but can't deal with this $n$
48461
How to prove the following limit?   $$\lim_{n \to \infty} (1+1/n)^n = e$$    I can only observe that the limit should be a very large number!   Thanks.
47213
$\lim n^n/(e^nn!)=0$ using Stirling approximation it is obvious. But can we do it without using Stirling approximation.  Now series with terms $x^n n^n/n!$ has ROC $1/e$. What we can say about behaviour of power series at endpoints?
34151
Given the fact that  $$\lim_{h\to 0} \frac{\sin(h)}{h}=1,$$ compute the following limit:  $$\lim_{h\to 0}\frac{\sin(x+h) - \sin(x)}{h}\ $$  How would I go about solving this problem? I have attempted to use trig identities (addition of angles) to try to simplify the problem, but it only seemed to make it worse...
7290
What is the value of the limit: $$\lim_{x \to 0} \frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})}$$ I think the answer should be $1$, but one I overheard one of my teachers saying that it is actually undefined.
15280
Can anyone help me with this limit:  $$ \lim_{n \to \infty}{\left(e^{\frac{1}{n}} - \frac{2}{n}\right)^n}$$  I think that I should expand $e^{\frac{1}{n}}$ as the sequence $1 + \frac{1}{n} + \frac{1}{n^{2}2!} + \frac{1}{n^{3}3!} + \dots $. But I can't see how it helps me.  Edit: or maybe use logarithm?