(Proof necessary)   $$\lim_{n \to \infty}  \frac{(n!)^{\frac{1}{n}}}{n}$$  I don't have an answer yet, but I know it exists, and is less than $1$.  Edit. Winther's answer is the most correct I don't understand how he is jumping from (log(n!) - nlog( n )) to it equal to the Sum from k=1 to n of log(k/n). Don't presume, it's wrong, I need to go, and I'll keep looking at it when I get back  Any help is appreciated
12006
Show:$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \frac{1}{e}$$  So I can expand the numerator by geometric mean. Letting $C_{n}=\left(\ln(a_{1})+...+\ln(a_{n})\right)/n$. Let the numerator be called $a_{n}$ and the denominator be $b_{n}$ Is there a way to use this statement so that I could force the original sequence into the form of $1/\left(1+\frac{1}{n}\right)^n$
54106
$\lim_{n\to\infty}((\frac94)^n+(1+\frac1n)^{n^2})^{\frac1n}$  Here's what I did:  $\lim_{n\to\infty}((\frac94)^n+(1+\frac1n)^{n^2})^{\frac1n}\\ =(\lim(\frac94)^n+\lim((1+\frac1n)^{n})^n)^{\frac1n}\\ =(\lim(\frac94)^n+\lim e^n)^{\frac1n}\\$  Any hints on how to continue?  PS: no logs/integration/derivation because we haven't covered it.
83831
I dont get it, if I look at $$\lim \limits_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$$    It seems like it should be $1$ because $1 \over n$ is almost zero and $1^\infty$ is still $1$
120469
How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$ ? i know that its "1", but why? How can i calculate this?  Thank you very very much =)
116856
Sorry if this is a silly question but most books claim  $\log n &lt; n$ for $n \geq 1$ without providing any proof, saying it's too obvious. Could someone give me a rigorous proof? Is there some trick or method I forgot about?
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.
60294
How to solve the following limit question? $$\lim_{n\to\infty}\frac{1}{e^n}\Bigl(1+\frac1n\Bigr)^{n^2}$$  Thanks a lot.
39067
Is there a way of showing this limit without using L'Hopital's rule? Specifically using the substitution $u = -\log{x},$ with knowledge of the limit of $\lim_{x\to0+}\log{x}.$ Thank you.
76156
I need help with the following limit $$\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{\sqrt{kn}}$$  Thanks.
54750
Show that $$\sum\limits_{k=1}^n \frac{1}{k} \geq \log(n)$$ Use this to deduce that the series $$\sum\limits_{k=1}^\infty \frac{1}{k}$$ diverges.   Hint: Use the estimate $$\frac{1}{k} \geq \int_k^{k+1} \frac{1}{x}\,\mathrm{d}x$$   Note that all instances of "$\log$" mean the natural logarithm, the inverse of the exponential. After doing the integration, I've concluded that $\frac{1}{k} \geq \log(1+\frac{1}{k})&gt;-\log(k)$  I suspect I'm supposed to use the hint to arrive at the first inequality, and then since $\log(n)$ can become arbitrarily large as $n$ increases, so must the harmonic series (essentially the comparison test).  I'm having trouble making this jump. Hints are appreciated, but please no solutions.
68886
$\sum_{n=1}^\infty \log \frac{n+1}{n}$  $\sum_{n=1}^\infty \log \frac{n+1}{n}$ =  $\displaystyle\lim_{n \to{+}\infty}(\log2 - \log1)+(\log3-\log2)+...+(\log(n+1)-\log n)$=$\displaystyle\lim_{n \to{+}\infty}\log(n+1)\to \infty$. So, the series diverges.  Is my procedure correct?
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.
101463
Find the limit $$\lim_{n \to \infty}\left[\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)\right]$$   I take log and get $$\lim_{n \to \infty}\sum_{k=2}^{n} \log\left(1-\frac{1}{k^2}\right)$$
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.
88599
Find   $$\lim _{x\to \infty }x\Big(\ln\left(x+1\right)-\ln x\Big)$$  Here's how I do it:  $$x\Big(\ln(x+1)-\ln x\Big) = x\Bigg(\ln(x(1+\frac{1}{x}) - \ln x\Bigg)$$ $$x\Big(\ln x + \ln(1+\frac{1}{x}\Big)-\ln x =x\ln\Big(1+\frac{1}{x}\Big)$$ $$\ln\Bigg( \Big(1+\frac{1}{x}\Big)^x \Bigg) \rightarrow \ln 1^\infty = 0$$  What am I doing wrong? The answer is supposed to be $1$, but I get $0$.
109742
I have been trying to prove this by induction on $n\in \mathbb{N}$, but this approach seemed to get me nowhere. I have a suspicion it might be necessary to express $\log{n}$ as $\int_1^n 1/x\text{ }dx$, but I could not build a rigorous argument from that.  Any help greatly appreciated!
64033
I am trying to calculate the limit of  $$\lim_{n \to \infty} \frac{(2n-1)!}{(2n+1)!}$$  and I am stuck. I tried n division but that didn't work. Any help is appreciated