# Tag Info

30

A number ends in $m$ zeroes if and only if it is divisible by $10^m$. To be divisible by $10^m$ means to be divisible by $2^m$ and by $5^m$. Big factorials (nothing special about 85) are products of lots and lots of numbers. After a while, lots of those numbers are divisible by 2 (and powers of 2), and lots are divisible by 5 (and powers of 5), so the ...

12

First of all, you need to know that starting with $n = 5$, $n!$ will always have at least one zero in the end. $5! = 5\times4\times3\times2\times1 = 120$. In this example, the $5$ and one $2$ contribute to that zero. After this, at $10!= 3628800$, another zero is added because the $5$ from the '10' and another $2$ from the product $10!$ contribute to the ...

11

It should be pretty obvious why $10!$ and all higher factorial must all have at least one zero at the end: they're all divisible by $10$. $$10! = \mathbf{10} \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$$ If you think about it a bit more, it's also pretty obvious that $20!$ and any factorials above it must haveat least ...

10

The first thought that comes into my head is to write $$e - \sum_{k=0}^n \frac{1}{k!} = \sum_{k=n+1}^\infty \frac{1}{k!},$$ so that the given sum is equivalent to a double sum: \begin{align*} \sum_{n=1}^\infty \sum_{k=n+1}^\infty \frac{1}{k!} &= \sum_{k=2}^\infty \sum_{n=2}^k \frac{1}{k!} \\ &= \sum_{k=2}^\infty \frac{1}{(k-2)!k} \\ &= ... 10 Note that \begin{align} \log a_n&=\log \sqrt{1!\sqrt{2!\cdots\sqrt{n!}}}=\frac{1}{2}\log 1! +\frac{1}{4}\log 2!+\cdots+\frac{1}{2^n}\log n! \\ &=\sum_{k=1}^n \frac{\log (k!)}{2^k}=\sum_{k=1}^n\frac{1}{2^k}\sum_{j=1}^k\log j= \sum_{k=1}^n \log k \Big(\sum_{j=k}^n \frac{1}{2^j}\Big). \end{align} Therefore, the sequence \log a_n, which is increasing, ... 6 One easy way is to observe in the binomial expansion of (1+1)^{2n} = \sum\limits_{k=0}^{2n} \binom{2n}{k}, the term \binom{2n}{n} is the largest one among all the 2n+1 terms of the form \binom{2n}{k}. This gives us a bound\frac{2^{2n}}{2n+1} \le \binom{2n}{n} \le 2^{2n} \quad\implies\quad \log 2 - \frac{\log(2n+1)}{2n} \le ...

5

Alternative computation: \begin{align*} a_n &= \frac{1}{2n} \log \binom{2n}{n} = \frac{1}{2n} \log \frac{(2n)!}{(n!)^2} \\ &= \frac{1}{2n} \sum_{k=1}^n \log \frac{n+k}{k} = \frac{1}{2n} \sum_{k=1}^n \log \Bigl( 1 + \frac{1}{k/n} \Bigr). \end{align*} Therefore, as $n \to \infty$, we get a Riemann sum: \begin{align*} \lim_{n \to \infty} a_n ... 5 \sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right) =\sum_{n=1}^\infty \int_0^1 \exp(u) \frac{(1-u)^{n}}{n!} du $$as everything is positive:$$ \sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right)= \int_0^1 \exp(u) \sum_{n=1}^\infty \frac{(1-u)^{n}}{n!} du \\= \int_0^1 \exp(u)(\exp(1-u) - 1) du = e - \int^1_0 \exp u du = 1 $$5 Interestingly, if we do a series expansion in Mathematica Series[(Gamma[n + 1]/Sqrt[Pi] (n/Exp[1])^-n)^6, {n, Infinity, 6}] We get$$8 n^3+4 n^2+n+\frac{1}{30}-\frac{11}{240 n}+\frac{79}{3360n^2}+\frac{3539}{201600 n^3}-\frac{9511}{403200 n^4}-\frac{10051}{716800n^5}+\frac{233934691}{6386688000n^6}+O\left(\frac{1}{n}\right)^{13/2}.$$This suggests that ... 5 "Higher order of growth" does not mean that n!\lt n^n but the stronger property that$$n!/n^n\to0.$$To prove that this property holds, note that n\geqslant2k for every 1\leqslant k\leqslant n/2 and n\geqslant k for every n/2\lt k\leqslant n, hence$$ n^n=\prod_{k=1}^{n}n\geqslant\prod_{1\leqslant k\leqslant n/2}(2k)\cdot\prod_{n/2\lt k\leqslant ...

4

Here is an estimate that gives a good approximation of $\binom{4n}{2n}$ in terms of $\binom{2n}{n}$. Using the identity $$(2n-1)!!=\frac{(2n)!}{2^nn!}\tag{1}$$ it is straightforward to show that $$\frac{\binom{4n}{2n}}{\binom{2n}{n}}=\frac{(4n-1)!!}{(2n-1)!!^2}\tag{2}$$ Notice that \begin{align} \frac{(2n-1)!!}{2^nn!} ... 4 Long time ago, needing an explicit formula for statistical thermodynamics, I was facing the same problem of the calculation of n! for 0<n<1. Starting from Ramanujan's approximation, what I did was to writen!\approx\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+x(n)}$$What can be easily established is that$$x(0)=\frac{1}{\pi ...

3

For $k \geqslant 2$, we have $$f(5^k) = \frac{5^k-1}{5-1} \equiv 1 \pmod{5},$$ so the multiple $$\frac{5^{k-1}-1}{4}\cdot 5$$ of $5$ is skipped. That produces the sequence $5,5+5^2 = 30, 5+5^2+5^3 = 155, 155+5^4 = 780,\dotsc$ of skipped multiples of $5$. There are more, for example $f(350) = 70+14+2 = 86$, so $85$ is skipped. Generally, if we call ...

3

You are right that $n^n$ grows faster than $n!$. You can see this question with answer (thanks to @JohnHabert for his comment above) for a proof which says exactly what you are saying. This is illustrated by just considering the first couple of numbers \begin{align} 1^1 = 1 \quad &\quad\quad 1! = 1\\ 2^2 = 4 \quad &\quad\quad 2! = 2\\ 3^3 = 27 ... 2 If n\in \mathbb{N}, Then n! = 1 \times 2 \times 3\times 4...............\times (n-2)\times (n-1)\times n Now n\geq n and n>(n-1) and n>(n-2) and n>(n-3)......... n>4\;\;,n>3\;\;,n>2\;\;,n>1 So n\cdot n\cdot n\cdot ............\cdot n(n-\bf{times})\geq n\cdot (n-1)\cdot (n-2)....3\cdot 2 \cdot 1 So n^n\geq n! 2 There can't be any multiplicative formula of the sort you describe for 4n\choose 2n in terms of 2n\choose n, because there are prime factors of the former that aren't factors of the latter. By Bertrand's Postulate there's a prime number between k=2n and 2k(=4n); that prime will be a factor of 4n\choose 2n, but can't be a factor of 2n\choose n ... 2 There is an important self-similar sequence that goes: 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, where I've grouped terms into bigger blocks. The "pattern" that makes this self similar is "=, =, =, =, +1" where "=" means repeat the working block unit and "+1" means to increment the block unit by adding one to the last ... 2 As suggested in several comments, the simplest form of Stirling approximation for n! is\sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}$$(http://en.wikipedia.org/wiki/Stirling%27s_approximation) Take the logarithms and develop first$$\log\left(2n!\right) - 2\log\left(n!\right)$$which results to be (2 n+1) \log (2). The remaining is obvious. If I may, ... 2 Setting$$ a_n=\frac{n^n}{n!}, $$we have$$ \frac{a_{n+1}}{a_n}=\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}=\frac{(n+1)^n\cdot n!}{n!\cdot n^n}=\frac{(n+1)^n}{n^n}=\left(1+\frac{1}{n}\right)^n \quad \forall n. $$Since$$ \lim_n\frac{a_{n+1}}{a_n}=\lim_n\left(1+\frac{1}{n}\right)^n=e>2, $$there is an N \in \mathbb{N} such that$$ ...

1

We show that $\lim_{x \to \infty} \frac{\sum{1 \leq i \leq n}log(i)}{n log(a)} = \infty$. Indeed, $\sum_{1 \leq i \leq n}log(i) > \sum_{n/2 \leq i \leq n}log(i)$. Note that for all $i \geq n/2$, we have $log(i) \geq log(n/2) = log(n)-1$. Hence, we have $\sum_{n/2 \leq i \leq n}log(i) \geq \frac{n}{2}log(n) - \frac{n}{2}$. Therefore, $\sum_{1 \leq i \leq ... 1 Using the limit formula: http://www.sms.edu.pk/journals/jprm/jprmvol8/01.pdf. Using the integral formula: http://www.math.unl.edu/~sdunbar1/ProbabilityTheory/Lessons/StirlingsFormula/GammaFunction/gammafunction.pdf. 1 If you wish to make an inverse of the factorial for a fractional value, try starting with one of the aproximations here and solving for$x\$. The two formulas are $$x!\approx\sqrt{2\pi}x^xe^{-x}\sqrt{x+\frac{1}{6}+\frac{1}{72x}-\frac{31}{6480x^2}-\frac{139}{155520x^3}+\frac{9871}{6531840x^4}}\\ ... 1 The matrix of Eulerian-numbers might help here. There are two (only slightly) different definitions; let's use that of the matrix which begins with$$ E=\small \begin{bmatrix} 1 & . & . & . & . & . \\ 1 & 0 & . & . & . & . \\ 1 & 1 & 0 & . & . & . \\ 1 & 4 & 1 & 0 & . & . ...

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