I need to calculate $$\lim_{n\rightarrow \infty} \left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$$ My try:  When $n!$ is large we have $n!\approx(\frac{n}{e})^n\sqrt {2\pi n}$ (Stirling approximation) $$\lim_{n\rightarrow \infty} \left(\frac{n!}{n^n}\right)^{\frac{1}{n}}=\lim_{n\rightarrow \infty}\frac{\left((\frac{n}{e})^n\sqrt {2\pi n}\right)^{\frac{1}{n}}}{n}$$ Simplifying we get, $$\frac{1}{e}\lim_{n\rightarrow \infty} \left(\sqrt{2\pi n}\right)^{\frac{1}{n}}$$  I am stuck here. I don't know how to proceed further.
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.
29097
Calculate below limit $$\lim_{n\to\infty}\left(\sum_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$$
121955
During my study I find a form for gamma function it was $\Gamma (x) = \lim_{n\to\infty} \frac{n! n^{x-1}}{x(x-1)(x-2)........(x+n-1)}$ And then by simplify this limit I get  $$\lim_{n\to\infty} \frac{n^{x-1}}{{x+n-1\choose n}}$$      then I use Stirling's approximation and evaluate the limit to $\ (x-1)! * e^{x-1} * (\frac{n}{n+x-1})^{x+n-1{\over2}}$$ What next I should do to get Gamma function from this limit ?
21873
Computing $\displaystyle \lim_{n\rightarrow \infty}\left(\frac{1^n+2^n+\cdots \cdots +n^n}{n^n}\right)$  Attempt: $\displaystyle \lim_{n\rightarrow \infty}\bigg[\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots \cdots +\left(\frac{n-1}{n}\right)^n+\left(\frac{n}{n}\right)^n\bigg] = 1$  Is my answer right? If not, then could someone help me? Thanks.
20154
Could be the following limit computed without using Stirling's approximation formula?  $$\lim_{n\to\infty} \frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}}$$  I know that the limit is $e$, but I'm looking for some alternative ways that doesn't require to resort  to the use of Stirling's approximation. I really appreciate any support at this limit. Thanks.
104524
As part of homework I'm trying to find the limit of $\sqrt[n]{{kn \choose n}}$ (with $k\in\mathbb{N}$ a given parameter).  I've seen , on which the only answer suggests using Stirling's approximation, which I've never learned and I can't seem to understand.  Trying to solve this myself, I first tried finding the ratio directly, which gave me $$\frac{{kn+k \choose n+1}}{{kn \choose n}}=\frac{\frac{\left(kn+k\right)!}{\left(n+1\right)!\left(kn+k-n-1\right)!}}{\frac{kn!}{n!\left(kn-n\right)!}}=\frac{\left(kn+k\right)!n!\left(kn-n\right)!}{\left(n+1\right)!\left(kn+k-n-1\right)!kn!}  =\frac{\left(kn+k-n\right)\left(kn+k-n+1\right)\cdots\left(kn+k\right)}{\left(n+1\right)\left(kn-n+1\right)\cdots\left(kn-n\right)} $$ That I'm not really sure how to use.  So instead I tried finding something that I might be able to say is bigger and smaller to use squeeze on, which gave me: $${kn \choose n}=\frac{kn!}{n!\left(kn-n\right)!}=\frac{\left(kn-n+1\right)\cdots\left(kn\right)}{n!}\leq k^{n}\frac{n^{n}}{n!}$$ And then showing that  $$ \frac{\frac{\left(n+1\right)^{n+1}}{\left(n+1\right)!}}{\frac{n^{n}}{n!}}=\frac{\left(n+1\right)^{n+1}n!}{n^{n}\left(n+1\right)!}=\frac{\left(n+1\right)^{n}\left(n+1\right)}{n^{n}\left(n+1\right)}=\left(1+\frac{1}{n}\right)^{n}\to\epsilon$$ gives the larger limit of   $$\sqrt[n]{{kn \choose n}}\leq\sqrt[n]{k^{n}\frac{n^{n}}{n!}}=\sqrt[n]{k^{n}}\sqrt[n]{\frac{n^{n}}{n!}}=k\cdot\sqrt[n]{\frac{n^{n}}{n!}}\to k\cdot e  $$  But it's obviously not the limit itself since for i.e. $k=1$ the limit is 1 and $k=2$ the limit is 4.  So I'm pretty much stuck, any hints on how I might be able to solve it?
89005
I have to prove that $$\lim_{n\to\infty} \left(1+n^2\right)^{\frac1n}$$ converges and give its limit. What I did was the following: $$\lim_{n\to\infty} \left(1+n^2\right)^\frac{1}{n} = \lim_{n\to\infty}(n^2)^{\frac1n}\lim_{n\to\infty}\left(\frac1{n^2} + 1\right)^\frac1n= \lim_{n\to\infty}(n^2)^{\frac1n}$$ Now that is where I am stuck - how to I continue with $\lim_{n\rightarrow\infty}(n^2)^{\frac{1}{n}}$? I really don't have any ideas left... Thank you very much! FunkyPeanut
101463
Find the limit $$\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)$$
110152
I need to prove that $$\lim_{n\rightarrow \infty} \frac{n^k}{2^n}=0$$ where $k\in \mathbb{N}$. All I can think of is to use something like L'Hopital's rule but I suppose there must be a another simpler way. I would much appreciate if someone could give me a hint. Thanks
109362
Evaluate the limit      $$ \lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} $$   My Attempt:  $$ \lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} = \lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^{n^2}\cdot \frac{1}{e^n} $$  How can I solve the problem from this point?
72717
Now this exercise looks like I need to use $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n=e$$ somehow, but I'm unsure of how. Bernoulli would give me $$\left(1+\frac{1}{n^2}\right)^n &gt; 1+n\cdot\frac{1}{n^2}=1+\frac{1}{n}$$ for which the limit is $1$, but I'm not sure how that helps. Can anybody give me a tip on how to solve this elegantly?
15280
Can anyone help me with this limit:  $$ \lim_{n \to \infty}{\left(e^{\frac{1}{n}} - \frac{2}{n}\right)^n}$$  I think that I should expand $e^{\frac{1}{n}}$ as the sequence $1 + \frac{1}{n} + \frac{1}{n^{2}2!} + \frac{1}{n^{3}3!} + \dots $. But I can't see how it helps me.  Edit: or maybe use logarithm?
73834
Find the limit of: $$\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n$$   I'm pretty sure it goes to zero since $(n+1)^n &gt; n^n$ but when I input large numbers it goes to $0.36$.  Also, when factoring: $$n^{1/n}\left(\frac{1}{1+\frac1n}\right)^n$$ it looks like it goes to $1$.   So how can I find this limit?
95156
Determine the following limit $$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{(n+1)(n+2)\ldots (n+n)} $$ Idea #1 the nth root could be rewritten as $$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{\frac{(n+n)!}{n!}} $$ Noticing that $n^{-1}$ is an infinitesimal, if the n-th root were bounded, the limit would be $0$.  But hoping that $\sqrt[n]{\frac{(n+n)!}{n!}}$ is bounded is a pipe dream, this clearly tends to $\infty$.  Idea #2 $$\lim_{n\to\infty}\ln\left (\frac{1}{n}[(n+1)\ldots (n+n)]^\frac{1}{n}\right ) = \lim_{n\to\infty}\left [-\ln n+\frac{1}{n}\left (\ln (n+1)+\ln (n+2)+\ldots +\ln (n+n) \right )\right ]$$ but trying to bring it to a common denominator doesn't help either, what I would have is $$\lim_{n\to\infty}\frac{1}{n} \left [-n\ln n+\ln (n+1)+\ldots +\ln 2+\ln n\right ]$$ Which doesn't bring me any closer to simplification. The solution of this must something entirely different. I would like a hint/point me in the direction I should go.
118166
Calculate $$\lim_{n\rightarrow +\infty}\binom{2n} n$$ without use Stirling's Formula.  Any suggestions please?
1860
How to show that  $$\lim_{n\to\infty}\left[\sum^n_{k=1}\frac{1}{k}-\ln(n)\right]=0.5772\ldots$$  No clue at all. Need help!  Appreciated!
52587
Find $$\lim_{n \to \infty} \frac {1}{2^n} \sum_{k=1}^n \frac{1}{\sqrt{k}} \binom{n}{k}.$$  First time I thought about Stirling's approximation but didn't get anything by applying it. I would also think about a Riemann Sum, but no idea how to rewrite...  The answer is $0$.