I found on some online sites that $$  \lim_{ n \to \infty}  \sqrt[n]{n!}= \infty $$ But I did not undertand ow they get it. My attempt: $$  \lim_{ n \to \infty}  \sqrt[n]{n!}= \lim_{ n \to \infty}  e^{\ln[ (n!)^{\frac1n}]} = \lim_{ n \to \infty}   e^{\frac1n \ln(n!)} = \lim_{ n \to \infty}   e^{ \frac{\ln(n)}{n} +\frac{\ln(n-1)}{n}+ \cdots } =   e^{ 0 +0+ \cdots }=   e^0 =1 $$
12869
I am playing around with the root/ratio test to practice with series. I just showed that $\sum \frac{1}{n!}$ converges by using the ratio test. I decided to see how things would go with the root test and I got stuck at something that I can't find on google. Right away while running the root test I encountered $\lim_{x \to +\infty}\left(\frac{1}{n!}\right)^\frac{1}{n}$. I have made the claim that $\left(n!\right)^\frac{1}{n}\geq 1 \hspace{3mm}\forall n\in \mathbb{N}.$ I have begun the proof and I think it is on the right track but some verification would be nice.  Proof: We can see that for n=1, this obviously holds, $\left(1!\right)^1\geq 1.$ Now suppose that this is the case for some $n=k$, then we have $\left((k+1)!\right)^{\frac{1}{k+1}}= \left((k+1)k!\right)^\frac{1}{k+1}$. It is at this point that I start to have a little trouble. Can somebody give me a push in the right direction? Thanks!
59359
Does $\displaystyle\lim_{n\to \infty}\ \frac{1+\sqrt[2]{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}=1$?   Thanks a lot for your time and help.
110771
I am trying to show that $\lim_{n\to \infty}{\frac{2n\choose n}{4^n}}=0$. I found that using stirling's approximation, I can get: $$ \lim_{n\to \infty}{\frac{2n\choose n}{4^n}}= \lim_{n\to \infty}{\frac{(2n)!}{n!^2*2^{2n}}}= \lim_{n\to \infty}{\frac{(\frac{2n}{e})^{2n}\sqrt{2\pi (2n)}}{((\frac{n}{e})^n\sqrt{2\pi n})^2*2^{2n}}} =\lim_{n\to \infty}{\frac{2\sqrt{\pi n}}{2\pi n}} =0 $$  But this seems inelegant where there should be a more elegant, combinatorial proof. Is there an easier way?
2943
I'm interested in the exact barrier of $$\ln(n!)= \Theta(n\ln(n))$$  and if it even exists.  This means, there is a $c_1, c_2$ , so that $$\ln(n!) \le  \Theta(n\ln(n) c_1$$  $$\ln(n!) \ge  \Theta(n\ln(n) c_2$$  or  $$0 \le \lim\limits_{n \to \infty}\frac{\ln(n!)}{n\ln(n)} \ge \infty$$  I could use de l'Hospital for the limit fraction, since it's $\frac{\infty}{\infty}$ but the derivate of $\ln(n!)$ causes problems. Also I don't know how I could prove the two statements above or how to prove them wrong. Any help is appreciated.
24947
When $\lim_{n\to\infty} x_n = a$, and $\lim_{n\to\infty} y_n = b$, find the limit, $$\lim_{n\to\infty} \frac{x_1 y_n + x_2 y_{n-1} + \cdots + x_n y_1}{n}.$$ Thank you for your help in advance.
31094
$$\lim_{n\to \infty}2^{\frac{1}{n}}$$   Basically looking at $$\lim_{n\to \infty}2^{\frac{1}{n}}=2^{\lim_{n\to \infty}{\frac{1}{n}}}=2^0=1$$  But if I use $$\lim_{n\to \infty}2^{\frac{1}{n}}=\lim_{n\to \infty}e^{\ln\left(2^{\frac{1}{n}}\right)}=\lim_{n\to \infty}e^{{\frac{1}{n}}\cdot{\ln(2)}}=\lim_{n\to \infty}e^{\frac{1}{n}}\cdot\lim_{n\to \infty}e^{\ln(2)}=1\cdot2=2$$  Where did I got it wrong?
117850
When calculating the limit $L=\big( 1 - \frac 1 {\sqrt n}\big)^n$, $n\to\infty$, what allows me to do the following:  $$ L=\lim \left(\left(1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n} $$  As the term inside the outer parenthesis goes to $e^{-1}$, we have $L=\lim e^{-\sqrt n}=0$.  It's like we're distributing the parenthesis somehow:  $$\lim \left(\left(1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}=\lim \left(\lim\left (1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}=\lim e^{-\sqrt n}$$  The question is: Why can we do this? Which property are we using?
53954
How can we compute the following limit:     $$\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$$   Mathematica gives the answer $\sqrt{e}$. However, I do not know to do it.
53994
Problem statement: Prove that $\frac{1*3*5*...*(2n-1)}{2*4*6*...(2n)}\leq \frac{1}{\sqrt{2n+1}}$ and that there exists a limit when $n \to \infty $. , $n\in \mathbb{N}$   My progress  LHS is equivalent to $\frac{(2n-1)!}{(2n)!}=\frac{(2n-1)(2n-2)(2n-3)\cdot ....}{(2n)(2n-1)(2n-2)\cdot ....}=\frac{1}{2n}$ So we can rewrite our inequality as:  $\frac{1}{2n}\leq \frac{1}{\sqrt{2n+1}}$ Let's use induction:  For $n=1$ it is obviously true. Assume $n=k$ is correct and show that $n=k+1$ holds.  $\frac{1}{2k+2}\leq \frac{1}{\sqrt{2k+3}}\Leftrightarrow 2k+2\geq\sqrt{2k+3}\Leftrightarrow 4(k+\frac{3}{4})^2-\frac{5}{4}$ after squaring and completing the square. And this does not hold for all $n$  About convergence: Is it not enough to check that $\lim_{n \to \infty}\frac{1}{2n}=\infty$ and conclude that it does not converge?
62524
strong textFind $\lim_\limits{n\to \infty}\left({1\over \sqrt{n^2+1}}+{1\over \sqrt{n^2+2}}+\cdots+{1\over \sqrt{n^2+n}}\right)$. I do know it is bounded by $1$. I tried using the sandwich rule with no success. How can I solve it?
2244
I'm trying to find  $$\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}} .$$   I tried couple of methods: Stolz, Squeeze, D'Alambert  Thanks!  Edit: I can't use Stirling.
21544
Help with Limit: $$\lim_{n \to \infty} \frac{(2 n)! e^n (n)^n}{n! (2 n)^{2 n}}$$  EDIT: Posted a similar question just yesterday but bad formating from my part dropped a term, and lead me to strange places. Honestly, I am just plain stuck, I have been hitting my head against it for $2$ days straight. I know the solution should be $2^{-1/2}$ but...  Help would be appreciated. If anyone has a Wolfram Alpha PRO account I am sure that thing churns out the solution as it is something quite basic I just don't see it and its driving me crazy.
106235
Evaluate the limit : $$\lim_{n \to \infty } {1 \over {\sqrt n }}\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)$$  I can use the sandwich principle, certain convergence criteria, Cesaro mean theorem, limit arithmetic.. things around this area.  Any help would be greatly appreciated, thanks!  Sorry for not elaborating more at the beginning, rookie first-post mistake I suppose. :)
32984
$$\lim_{n \to \infty} n\bigg(1-\dfrac{1}{e}\bigg(1+\dfrac{1}{n}\bigg)^{n} \bigg)$$   If I write expansion of $\bigg(1+\dfrac{1}{n}\bigg)^{n}$ it was equal to expansion of $e$ so $n-n=0$. Is limit is zero ?  Edit: Uploading screen shot of my response. They marked it correct and I don't know if answer key is wrong or not ? Please can anyone say for sure like with $100$ percent  surety if answer given is wrong.
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))=$$=n(ln(n\sqrt{{2n\pi n})})$ and that would imply that $\lim_{n\rightarrow\infty}a_{n}=exp(n\sqrt{2\pi n})^n$ what does not seem good... Where am I doing mistake, is it wrong way of acting? Any hints are desired, thank you in advance.
23066
How to compute the following limit?  $\lim\limits_{n\to\infty} n&#183;(\sqrt[n]{a}-1)$  We know that it's got something to do with $\ln$ or $\exp$.  We know that $\lim\limits_{n\to\infty} \sqrt[n]{a} = 1$ but it seems to not to be true that therefore $\lim\limits_{n\to\infty} n&#183;(\sqrt[n]{a}-1) = \lim\limits_{n\to\infty} n&#183;(1-1) = 0$.  What we know is  $\lim\limits_{n\to\infty} (1-\frac{1}{n})^n = \lim\limits_{n\to\infty} (1 + \frac{1}{n})^n \lim\limits_{n\to\infty} (1-\frac{1}{n})^{n+1} = e$  and  $\lim\limits_{n\to\infty} (1+\frac{x}{n})^n = \exp(x)$
70203
I need to check if $$\lim_{n \to \infty} \frac{n}{ \sqrt [n]{n!}}$$ converges or not. Additionally, I wanted to show that the sequence is monotonically increasing in n and so limit exists. Any help is appreciated. I had tried taking log and manipulating the sequence but I could not prove monotonicity this way.