I tried take limit n goes to infinity. Wolfram solved the limit and $(n^2)!$ has greater growth rate but there is not step by step solution. I think I can simplify $(n!)^n$ to $(n^n)^n$ But how can I prove $(n^2)! &gt; (n^n)^n$
123037
It is well known that $(n!)^2$ divides $(2n)!$.  Does it follow that $(n!)^3$ divides $(3n)!$ and so on up to $(n!)^n$ dividing $(n^2)!$?  If yes or no, could you provide the details behind the argument?  As a quick test, I tried $n=3$ and $n=4$ which worked.  Thanks,  -Larry
18135
How do I prove that $3^n &gt; n^2$ with mathematical induction? I thought I had the correct answer but my teacher says its wrong.  I let $n=1$ for the initial case and it works. I then assumed $n=k$ works and went onto the $n=k+1$ case, but after that it goes a little awry. Thanks in advance!
86681
Find $\lim\limits_{n\rightarrow \infty} \frac{n^t}{e^n}$ where $t$ is a positive real number.   My answer is that the expression is indeterminate ($\frac{\infty}{\infty}$) but it will tend to $0$ because as $n$ gets large, the exponential term will dominate the polynonial term in the expression, so the expression will tend to $0$ as $n$ gets large. This answer is also the same as the solution provided in my text.  But I am wondering what is the actual workings to solve this? I'd know to because I want to be familiar with all the tricks to solve limits where the expression is indeterminate.
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.
93160
I need to find the limit, not sure what to do. $\lim_{x \to \infty} \sqrt{x^2 +ax} - \sqrt{x^2 +bx}$  I am pretty sure I have to divide by the largest degree which is x^2 but that gets me some weird numbers that don't seem to help.
66113
Let $\ a_n=\sqrt{n+1}-\sqrt{n}$. I have to show that $\lim_{n\to \infty}a_{n}=0$.     How should I start? Do I have to use any theorem?
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.
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.
2770
Show that if $n&gt;2$, then $(n!)^2&gt;n^n$.     My work: I tried to apply induction. So, at the induction step, I need to prove, $n^n&gt;(n+1)^{n-1}$ Here, I tried to use induction again without any luck. I also took log of both sides, but I did not get anything. Please help!
7100
$f(n) = n + (\log n)^{2} , g(n) = n + \log(n^{2} ).$  Now so far I've done $g(n) = n+ 2\log(n)$  and then I think since its the same change to both, I can remove $n$.  leaving me with $f(n) = (\log n)^{2}$ and $g(n) = 2\log(n)$ now if I do $f(n)/g(n)$ I  get  $((\log n)^{2}) / (2\log n)$   and since the numerator is growing exponentially and the bottom is growing linearly? As we approach positive infinity the function approaches positive infinity. Thus $f(n)$ is $\omega$ of $g(n)$?  I don't know if this proof is done right, if its what is a better alternative? I am still new to this and learning!
60296
I am a CS undergrad and I'm studying for the finals in college and I saw this question in an exercise list:     Prove, using mathematical induction, that $2^n &gt; n^2$ for all integer n greater than $4$   So I started:  Base case: $n = 5$ (the problem states "$n$ greater than $4$", so let's pick the first integer that matches)  $2^5 &gt; 5^2 \implies 32 &gt; 25$ - ok!  Now, Inductive Step:  $2^{n+1} &gt; (n+1)^2$  now expanding  $2 * 2^n &gt; n^2 + 2n + 1$  Well, I can't go on, I don't know how to finish this proof. Could anyone help me with this one?  Also, I know that the statement (inductive hypothesis) is true. Simply, I'll have a number to an exponent $n$ twice, next to $n$ squared and $2n$ (which could be seen as quadratic and linear functions) which "grows" much lower than the exponential one. It could be easily "proved by example" with any $n$ greater than $4$.  I want to know how to prove it algebraically , so my professor does not freak out.  thanks in advance.
75653
If $0 &lt; r &lt; 1$ show that $r^n$ goes to $0$ as $n \to 1$.  If $|r| &lt; 1$ then  $r^2  &lt; r$  similarly $r^4 &lt; r^3 &lt; r^2 &lt; r$  so $r^n$ as $n \to +\infty$ will be equal to $1$  how can i finish this problem?
107601
Find the limit following:  $$L=\lim \limits_{n\to \infty}\sqrt[n]{x^n+y^n+z^n}$$  With $x,\: y\: z\in R$  P.S  I think this limit result is $L=max\left\{x,\: y\: z \right\}$. But i'm not find it, so expect people to help me find out the results by some solution  Extend this limit:  $$S=\lim \limits_{n\to \infty}\sqrt[n]{\sum_{i}^{m}a_{i}}$$ with $a_i\in R,\: i=1,\: m$
34769
These are all from Sipser's book, second edition. I was just hoping someone could verify/explain those that are more difficult for me.  $2n = O(n)$: true  $n^2 = O(n)$: false  $n^2 = O(n\log^2 n)$: I say false, because $\log^2 n$ grows more slowly than $n$  $n\log n = O(n^2$): true because $n$ grows more quickly than $\log^2n$  $3^n = 2^{O(n)}$: I wasn't sure what this notation meant. Any clarification here?  $2^{2^n} = O(2^{2^n})$: I would say true, but I'm not entirely sure why.  $n = o(2n)$: false, because $o(2n) = o(n)$ and $n$ grows as quickly as itself.  $2n = o(n^2)$: true, $n^2$ is always more than $2n$ for some constant.  $2^n = o(3^n)$: true  $1 = o(n)$: I'm not really sure.  $n = o(\log n)$: false, $n$ is always larger.  $1 = o(1/n)$: true, because 1 is always larger.
49267
The limit to solve is $$\lim_{n\to\infty}\left(\frac{n^2+2}{2n^2+1}\right)^{n^2}$$ I tried to use L'Hôpital's rule but the derivatives are quite complex.
29302
How would you go about proving $n! &lt; n^n$ using a mathematical induction proof?  I understand how to solve inductive proofs with = but I'm getting a bit lost in this example.   Any help is much appreciated.
78161
I'm trying to find the limit of $a_n = \left(1-\frac{1}{n^2}\right)^n$ for $n \rightarrow \infty$.  It seems that the limit is $1$, since $a_n = 0.999...$ for large $n$. The presentation $a_n = \frac{(n^2-1)^n}{n^{2n}}$ and expanding was my first idea, but I couldn't get the result from there. Any ideas?