I want to calculate   $  \lim_{n \to \infty }{\frac{{[n(n+1)(n+2)...(2n-1)]}^\frac1n}n} $  using   $$\int_0^1 f(x)\,dx=\lim_{n\to\infty}\frac1n \sum_{k=1}^n f\left(\frac{k}n\right)$$  I know I have to convert $\frac nk$ to $x$, but I am confused since all the factors are multiplied together. Should I use $\log$?
55350
Evaluate $$\lim_{n \rightarrow \infty~} \dfrac {[(n+1)(n+2)\cdots(n+n)]^{\dfrac {1}{n}}}{n}$$  Attempt: Let $$y=\lim_{n \rightarrow \infty} \dfrac {[(n+1)(n+2)\cdots(n+n)]^{\dfrac {1}{n}}}{n}$$  $$\implies \log y = \lim_{n \rightarrow \infty} \dfrac {1} {n} [\log (n+1) +\cdots+log(n+n)-log(n)] $$  How do I move forward?  Thank you very much for your help.
97971
$$\lim_ {n\to \infty}\ \frac{n}{n+1}=1$$   I'd write it like this:  $$\lim_ {n\to \infty}\ \frac{n}{n}+\frac{n}{1}=\lim_ {n\to \infty}\ 1+n=\infty$$  What am I missing?
103605
Evaluating the limit $\displaystyle \lim_{n\to\infty} \left[  \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$  I have a question about the following solution:  We may write it in the form:  $$ \frac{1}{n} \left[ \frac{1}{1+(\frac{1}{n})^2} + \ldots + \frac{1}{1+(\frac{n}{n})^2} \right] $$  Somehow I need to figure out that the limit is actually the Riemann sum of $\frac{1}{1 + x^2}$ on $[0,1]$ for $\pi = 0 &lt; \frac{1}{n} &lt; \ldots &lt; \frac{n}{n}$.  Can you explain to me to reach this conclusion?
86379
I have to solve: $\lim\limits_{x\to \infty} \frac{\ln(x+1)}{\ln(x)}$. Can you give me any hints to go? Thanks a lot!
99286
The limit is    $$\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$$  The problem is I don't know if I can calculate it normally like with a change of variables or not. Keep in mind that I'm not allowed to use L'Hôpital's rule nor the $\mathcal O$-notation.
119125
Evaluate the value of $$\lim_{n \to \infty}\frac{x_n}{\prod_{i=1}^{n-1} x_i}$$ where $x_n=x_{n-1}^2-2, (x_1=5).$  I've found that $x_n$ is an increasing sequence so that the limit becomes an indeterminate form of $(\frac{\infty}{\infty})$. I can't think of any other workable ideas though. Some hints are 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 $0$.
8168
I am trying to calculate the following limit  $$\lim_{N\to\infty} \frac{N!}{(N-k)!N^{k}}$$ where $k$ can be any number between $0$ and $N$.  I thought of the following:  If I take the logarithm of the expression then I get:  $$\lim_{N\to\infty} \left(\log(N!)-\log((N-k)!)-k\log N\right)$$    Using the Stirling formula this can be approximated as:  $$\lim_{N\to\infty} \left(N\log(N)-(N-k)\log(N-k)-k\log N\right)$$  Now there are two cases: If $k$ is $N$, then the second term vanishes and the remaining terms cancel. If $k$ is smaller than $N$, then I can drop the $k$ inside the second logarithm and all the terms cancel. So the limit $$\lim_{N\to\infty} \log\left(\frac{N!}{(N-k)!N^{k}}\right)=0$$ Which means that: $$\lim_{N\to\infty} \frac{N!}{(N-k)!N^{k}}=1$$ I don't know if this is mathematically rigorous. Would like some help. Thanks
121419
How to prove the following: $$ \lim_{n \to \infty} \int_0^1 \int_0^1 \cdots \int_0^1 \frac{x_1^2+x_2^2+ \cdots +x_n^2}{x_1+x_2+ \cdots +x_n} dx_1 dx_2 \cdots dx_n = \frac23 $$  I would really appreciate if you could help me!
108053
Find $\lim_{n\to\infty} ((n!)^{1/n})$. The question seemed rather simple at first, and then I realized I was not sure how to properly deal with this at all. My attempt: take the logarithm,  $$\lim_{n\to\infty} \ln((n!)^{1/n}) =  \lim_{n\to\infty} (1/n)\ln(n!) =  \lim_{n\to\infty} (\ln(n!)/n)$$ Applying L'hopital's rule:  $$\lim_{n\to\infty} [n! (-\gamma + \sum(1/k))]/n! = \lim_{n\to\infty} (-\gamma + \sum(1/k))=  \lim_{n\to\infty} (-(\lim(\sum(1/k) - \ln(n)) + \sum(1/k))  = \lim_{n\to\infty} (\ln(n) + \sum(1/k)-\sum(1/k)  = \lim_{n\to\infty} (\ln(n))$$ I proceeded to expand the $\ln(n)$ out into Maclaurin form $$\lim_{n\to\infty} (n + (n^2/2)+...) = \infty$$ Since I $\ln$'ed in the beginning, I proceeded to e the infinity  $$= e^\infty  = \infty$$  So am I write in how I approached this or am I just not on the right track? I know it diverges, I was just wanted to try my best to explicitly show it.
58267
How to compute this limit:  $$\lim_{n\to\infty}\left(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\cdots+\frac{n}{n^2+n^2}\right)$$  Please give me some hint.
44975
How may one calculate  $$\lim_{n\to\infty} \ \left(\left(\sum_{k=1}^{n} \frac{1}{3k-1}\right) - \frac{\ln n}{3}\right) \ ?$$
19015
$$\lim_{n\to \infty}\frac1{2n}\log\left({2n \choose n}\right)$$  Now, $\log(n!) = \Theta (n\log(n))$ so I think we could write,   $$\lim_{n\to\infty}\frac{1}{2n}\left(\log\left(2n!\right) - \log\left(n!^2\right)\right) = \frac{1}{2n}\left(\log\left(2n!\right) - 2\log\left(n!\right)\right) $$  $$\lim_{n\to\infty}\frac{1}{2n}\left(2n\log\left(2n\right) - 2n\log\left(n\right)\right) $$  $$\log\left(2n\right) - \log\left(n\right) = \log(2)$$  Is this legitimate? I feel this might be wrong since, the $\Theta$ notation conceals a constant factor.
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 $0$.
76156
I need help with the following limit $$\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{\sqrt{kn}}$$  Thanks.
29010
Given $\lim \limits_{x\to\infty}(1+\frac{1}{x})^{x}$, why can't you reduce it to $\lim \limits_{x\to\infty}(1+0)^{x}$, making the result "$1$"?  Obviously, it's wrong, as the true value is $e$.  Is it because the $\frac{1}{x}$ is still something even though it's really small?  Then why is $$\lim_{x\to\infty}\left(\frac{1}{x}\right) = 0\text{?}$$  What is the proper way of calculating the limit in this case?
45090
I need to find the following:  $$\lim_{x\to\infty} f(x)= \lim_{x \to \infty}\left (\frac{x}{x+1} \right )^x$$  I know that this limit = $\frac{1}{e}$ from plugging it into a calculator, but I have to prove it without using the fact that:  $$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}\left (\frac{x}{x+k} \right )^x=\frac{1}{e^k}$$  I started by exponentiating:  $$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}e^{\ln \left (\frac{x}{x+1} \right )^x}$$  and from here I've dropped the exponent $x$ in front of the $\ln$, and from here I'm getting stuck. Should I separate the $\ln$ like this?  $$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}e^{x(\ln(x)-\ln(x+1))}$$  This doesn't seem to be leading me down the right path, but I'm not sure how else to do it. Is there a way to apply L'Hopital? If so, how?