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5

The recurrence for $n!$ is $n! = n (n-1)!.$ Apply for $n=k+2.$

1

$$\frac{20!}{2^{40}}= \frac{20!}{4^{20}} = \left(\frac{1}{4}\right)\left(\frac{2}{4}\right)\left(\frac{3}{4}\right) \left(\frac{4}{4}\right)\left(\frac{5}{4}\right)\left(\frac{6}{4}\right)\left(\frac{7}{4}\right)\left(\frac{8}{4}\right)\left(\frac{9}{4}\right)\cdots \left(\frac{20}{4}\right)>\cdots$$ $$... 3 Hint: In general, for positive integer x,y we have: y!=y(y-1)(y-2) \cdot\cdot\cdot2\cdot1 x^y=xxx\cdot\cdot\cdot xx If x\geq y; then  x^y\ge y!  If x<y , ?! (Now this is the question!) Extra: If  \left\lceil{\frac{y}{2}}\right\rceil+1< x<y; then x^y\ge y!. (easy proof) Now, we have only the case  ... 3 "Just compute them and compare" is the only fully failsafe method. In most cases, however, estimating the logarithms of both of x^y and y! using Stirling's formula will yield a conclusive result without needing to compute the two values in high detail. 7 For a fixed x, y! will eventually be larger, as you can tell by using Stirling's formula. 2 My suggestion is to take logs of both sides, and for n! consider the integral of \log x. 0$$\frac{(n+2)!}{2} = (n+1)! + \sum_{i=1}^n i\,n!$$0 I have found the following formula:$$n!=\sum_{k=2}^{n+1}(-1)^{n+1-k}\binom{n+1}{k}\sum_{i=1}^{k-1}i^{n},\ n\in N.$$0 A reference for the complete asymptotic expansion of \prod_{k=0}^n \binom{n}{k}, together with the expansion itself, is given here: Prove that \prod_{k=1}^{\infty} \big\{(1+\frac1{k})^{k+\frac1{2}}\big/e\big\} = \dfrac{e}{\sqrt{2\pi}} 2 See "Inequalities for Gamma Function Ratios", G.J.O. Jameson, American Mathematical Monthly, December 2013, pp 936-940. This recent article gives elementary proofs of Gautschi-type inequalities. In particular, you get$$\left({x\over x+1/2}\right)^{1/2}\leq {\Gamma(x+1/2)\over\sqrt{x}\,\Gamma(x)}\leq 1.$$Now let x\to\infty. 1 Notice that the reciprocal of the expression under the limit is just the Euler's beta function (see wiki for details):$$ \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)} = \operatorname{B}\left(n, \frac{1}{2}\right) = \int_0^1 \frac{x^{n-1}}{\sqrt{1-x}} \mathrm{d}x $$Hence, using \Gamma(1/2)=\sqrt{\pi}:$$ ...

1

You are going around in circles, since you reduced a problem with two gammas to a problem with three gammas! Instead, the scientific way uses Stirling's expansion. Notice that the second (an in some ways easier) derivation is used for Gamma directly. (in any case, use the Stirling approximation for numerator and denominator, divide, take limits.

1

Here is another contender that does not use the Leibniz rule. The term that needs to be simplified here is $$\frac{n!}{(n-r)!} \sum_{k=0}^r {r\choose k} (-1)^k {n-r\choose i-r+k} x^{i-r+k} (1-x)^{n-i-k}$$ which is $$\frac{n!}{(n-r)!} x^i (1-x)^{n-i} \sum_{k=0}^r {r\choose k} (-1)^k \frac{(n-r)!}{(i-(r-k))!(n-i-k)!} \frac{1}{x^{r-k}} \frac{1}{(1-x)^k}$$ ...

1

You are on the right track. Your derivation is correct, but you need to to do more work to reach your destiny. Your last step $${{n}\choose{i}}\sum\limits_{k=0}^{r}{{r}\choose{k}}\dfrac{i!}{(i-k)!}x^{i-k}\dfrac{(n-i)!}{(n-i-r+k)!}(1-x)^{n-i-r+k}(-1)^k$$ can be simplified as $$= ... 0 Richard Guy, Unsolved Problems In Number Theory, 3rd edition, problem B23: Equal products of factorials, begins, Suppose that$$n!=a_1!a_2!\dots a_r!,\quad r\ge2,\quad a_1\ge a_2\ge\dots\ge a_r\ge2$$A trivial example is$$a_1=a_2!\dots a_r!-1,\quad n=a_2!\dots a_r!$$Dean Hickerson notes that the only nontrivial examples with n\le410 are ... 2 Simplifying your first attempt yields$$(1-c)^{-i} c^i \sum_{j=0}^n {n\choose j} {j\choose i} (1-c)^j u^j (1-u)^{n-j} .$$To evaluate the sum, observe that when we multiply two exponential generating functions of the sequences \{a_n\} and \{b_n\} we get that$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} \sum_{n\ge 0} b_n \frac{z^n}{n!} = \sum_{n\ge 0} ...

3

This has a nice proof if you interpret it in terms of probabilities. When $0<x<1$, $B^n_i(x)$ is the probability of $i$ successes from $n$ independent events if each event has a probability $x$ of success. Now, suppose that in order to succeed, you need to pass two (independent) stages: step 1 has a probability $u$ of success, and step 2 a ...

1

Suppose you want to find the number of ways to partition a set of n elements into k subsets or less, all of these non-empty and without the order of the partitions mattering. Then this is the same as $\sum_{j=1}^k S(n,j)$ where $S(n,j)$ denotes the number of way so partition a set of n elements into j subsets without the partitions being "labelled". the ...

1

When calculations involve large factorials, the use of Stirling approximation for n! is very convenient. In your case, you compare n! to n^(n/2). We can take logarithms of both sides and use the fact that Stirling approximation gives log(n!) close to n log(n) - n; the logarithm of the second term is simply n log(n) / 2. Using this approximation, you find ...

1

$300!=(300*1) * (299*2) * ... * (151*150)$, 150 pair products total $(300^{300})^{1/2}= 300^{150} = 300 * 300 * ... * 300$, 150 counts of 300 total none of pair product in the first line is smaller than $300$, so $300!$ is larger.

28

HINT: $(300^{300})^{1/2}=300^{150}$, so you’re comparing $$\underbrace{300\cdot300\cdot\ldots\cdot300}_{150\text{ factors}}$$ with $$300\cdot299\cdot\ldots\cdot1=\underbrace{(300\cdot1)\cdot(299\cdot2)\cdot\ldots\cdot(151\cdot150)}_{150\text{ factors}}\;.\tag{1}$$ Show that each of the parenthesized factors in $(1)$ is at least $300$.

1

As $\frac{(n!)!}{n!}=(n!-1)!$ for all n , there are infinitely many nontrivial triples (a,b,c) with $\frac{a!}{(a-b)!}=c!$ Choose a=n! , b=n!-n , c=n!-1

4

The problem isn't that $0!=1$, the problem is that the first term in the sequence doesn't match the pattern of the other terms. The sequence whose $n^{\text{th}}$ term is $\dfrac{(n-1)!}{n^2}$ looks like $$1,\ \dfrac{1}{4},\ \dfrac{2}{9},\ \dfrac{6}{16},\ \dots$$ whereas your sequence starts with $0$, not $1$.

1

For a positive series to converge, the $n$-th term in the series must go to $0$ i.e. $\displaystyle\lim_{n\to\infty} \frac{(2n^2-n+1)!}{3^{n^2+1}}=0$. But for $n\geq3$ we have $2n^2-n+1> n^2+1$. Hence $\displaystyle\frac{(2n^2-n+1)!}{3^{n^2+1}}=\frac{(2n^2-n+1)}{3}\frac{2n^2-n}{3}\cdots\frac{n^2-n}{3}\cdot{(n^2-n-1)!}>1$, showing that the $n$-th term ...

1

If $\displaystyle f(n)=2n^2-n+1,f(n+1)=2(n+1)^2-(n+1)+1$ So, $\displaystyle f(n+1)-f(n)=2(2n+1)-1=4n+1$ If $\displaystyle g(n)=n^2+1, g(n+1)=(n+1)^2+1\implies g(n+1)-g(n)=2n+1$ So, if $\displaystyle T_n=\frac{\{f(n)\}!}{3^{g(n)}},$ ...

1

Consider the $n$th entry $x_n=\dfrac{(a_n)!}{3^{b_n}}$ with $a_n=2n^2-n+1$ and $b_n=n^2+1$. The following facts hold for every $n\geqslant1$: $$a_n\geqslant n^2\qquad a_{n+1}\geqslant a_n+4n\qquad b_{n+1}-b_n\leqslant 3n$$ The two first facts yield $$... 2 Using the definition of the Exponential Series,$$e^x=\sum_{0\le r<\infty}\frac{x^r}{r!} \implies e^x-1=x+\frac{x^2}{2!}++\frac{x^3}{3!}+\cdots$$Taking nth power,$$ (e^x-1)^n=\left(x+\frac{x^2}{2!}++\frac{x^3}{3!}+\cdots\right)^n\text{Now, }(e^x-1)^n=e^{nx}-\binom n1e^{(n-1)x}+\binom n2e^{(n-2)x}-\cdots=\sum_{0\le ...

1

Your last step is not valid, because things do not repeat in this way (think about what happens when you reach a new power of 5, for instance). Since this is Project Euler I won't give a solution. However I will point out the (hopefully obvious) fact that you won't be able to solve this by considering all primes less than $10^{12}$, so you will need a ...

0

We should also note that the factorial function has a similar look to it as the sigma notation; as $$\frac{n(n+1)}{2}=1+2+3+...+n=\sum_{k=1}^nk$$ $$n!=1 \cdot 2 \cdot 3 \cdot ... \cdot n=\prod_{k+1}^nk$$

2

$\sum_{n=1}^{k} n = 1 +2+3+\ldots+k$. Is a nice notation for it. So $$1 + 2 + 3 + 4 + 5 = \sum_{n=1}^{5} n$$.

5

It is called the $n$th triangle number and it can be written as $\binom{n+1}2$.

0

That can be done with the formula $\frac{n^2+n}{2}$

5

$$x=\frac e5\implies\;\text{we have the series}\;\;\sum_{n=1}^\infty\frac{n!5^ne^n}{n^n5^n}=\sum_{n=1}\frac{n!e^n}{n^n}$$ and now you can use Stirling's Approximation $$n!\sim\frac{n^n}{e^n}\sqrt{2\pi n}$$ so our series behaves asimptotically (for large values of $\;n\;$) as the series $$\frac{n^n}{e^n}\sqrt{2\pi n}\frac{e^n}{n^n}=\sqrt{2\pi n}$$ and ...

1

You dont need to use permutation here because the ordering is not important . You will have to choose combination here . choosing $2$ red out of $8$ red = $_8C_2$ ways choosing $3$ black out of $7$ black = $_7C_3$ ways therefore total number of ways of doing (a)= $_8C_2 * _7C_3$

5

In this answer are three different proofs of this cancellation lemma: $$\sum_{j=k}^n(-1)^{n-j}\binom{n}{j}\binom{j}{k} =\left\{\begin{array}{} 1&\text{if }n=k\\ 0&\text{otherwise} \end{array}\right.\tag{1}$$ Inductively, we can see that any polynomial in $j$ of degree $m$ can be written uniquely as $$\sum_{k=0}^mc_k\binom{j}{k}\tag{2}$$ where the ...

1

As tenpercent said, since $n! \sim \sqrt{2\pi n} (n/e)^n$, \begin{align} \frac{3^n(2n)!}{n!(2n)^n} &\sim \frac{3^n\sqrt{2\pi 2n} (2n/e)^{2n}}{\sqrt{2\pi n} (n/e)^n(2n)^n}\\ &= \frac{3^n2^{2n}n^{2n}\sqrt{4\pi n} /e^{2n}}{\sqrt{2\pi n}2^n n^{2n}/e^n}\\ &= \frac{3^n2^{n}\sqrt{2} }{e^n}\\ &=\left( \frac{6}{e}\right)^n \sqrt{2} \end{align}

6

What's the problem with the ratio test?: $$\frac{a_{n+1}}{a_n}=\frac{(2n+2)!\color{red}{3^{n+1}}}{(n+1)!(\color{green}{2}(n+1))^{n+1}}\frac{n!(\color{green}{2}n)^n}{(2n)!\color{red}{3^n}}=\frac{(2n+1)\cdot3}{n+1}\left(\frac{n}{n+1}\right)^n\xrightarrow[n\to\infty]{}\frac6e>1$$ and thus...

0

Hint: use Stirling formula, $n! \simeq \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$

3

This identity also has an algebraic proof. Suppose the sum we are trying to evaluate is given by $$\sum_{k=0}^n {n\choose k} (-1)^k (n-k)^{n}.$$ Observe that when we multiply two exponential generating functions of the sequences $\{a_n\}$ and $\{b_n\}$ we get that $$A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} \sum_{n\ge 0} b_n \frac{z^n}{n!} = \sum_{n\ge ... 6 Let A:=\{ x_1,..,x_n \} and B=\{y_1,..,y_m \}. Lets count the number of onto functions f:A \to B. There are m^n functions from A to B. Lets count now the ones which are not onto: Define$$P_i= \{ f : A \to B |y_i \notin f(A) \}$$Then we need to figure out the cardinality of \cup_i P_i. By the inclusion exclusion principle$$|P_1 \cup ...

1

Your proof is nearly complete. For part (II), I would suggest that instead of concluding that $d=1$, conclude that the assumption for this part (that $d\nmid p$) contradicts the result that $d\mid 1$, therefore that this case leads to a contradiction and cannot occur. Then conclude that the result from part (I) is the only logical result, which is that ...

2

Since $(p-1)!-1$ is relatively prime to every integer smaller than $p,$ the only way the two quantities could fail to be relatively prime is if $p|(p-1)!-1.$ But $p|(p-1)!+1,$ and $p>2,$ so that's impossible.

-1

$n!> k^n$ if $n\ge ke$ try it for yourself with various combinations of $n$ and $k$ use $\log(n!)$ and $k \log(n)$ for large $n$

3

$\dfrac{8!}{5!}\cdot \dfrac{7!}{7!10!} = \dfrac{8!}{5!\cdot 8! \cdot 9\cdot10} = \dfrac{1}{120 \cdot90} = \dfrac{1}{10800}$.

3

$\dfrac{8!}{5!}\cdot ... 2 my steps:$ \dfrac{8!}{5!}\cdot\dfrac{7!}{7!\cdot10!} $7! and 7! cancel out:$\dfrac{8!}{5!} \cdot \dfrac{1}{10!}$next:$ \dfrac{8!}{5!} \cdot \dfrac{1}{10 \cdot 9 \cdot 8!} $Here, 8! cross-cancel :$\dfrac{1}{5!} \cdot \dfrac{1}{10 \cdot 9} $next:$ \dfrac{1}{10 \cdot 9 \cdot 5!}$which is$\dfrac{1}{ 10 \cdot 9 \cdot 5 \cdot4 \cdot3 \cdot2\cdot1}$... 1$n!$can be expressed as a polynomial of$n$: $$n! = \prod_{i=0}^n (n-i) = n^{n+1} + \cdots$$ This shows that$n!$is a$n+1$-degree polynomial of$n$. Then, the greatest power of$n$that you can find in this polynomial is$n^{n+1}$. The greatest power that you can obtain from$n!n!$is$n^{n+1}n^{n+1} = n^{2n+2}$with coefficient$1$. Also, the greatest ... 0 Hint:$U_n$is decreasing, limit exist.$U_n U_{n+1}=\frac{1}{n+1}$, let$n$go to infinity. 0 $$\sum_{i=1}^\infty \frac{1}{\sqrt t!}$$ Ratio Test $$\lim_{x \to \infty} |\frac{a_{n+1}}{a_n}| = L$$ $$|\frac{1}{\sqrt {(t+1)!}} \cdot \frac{\sqrt t!}{1}| = \frac{\sqrt{t!}}{\sqrt{(t+1)!}}$$ From Wolfram, $$\lim_{t \to \infty} \frac{\sqrt{t!}}{\sqrt{(t+1)!}}= 0$$ Since$L= 0 <1\$, $$\sum_{i=1}^\infty \frac{1}{\sqrt t!}$$ converges absolutely ...

3

Observe: $$\frac{a_{n+1}}{a_n}=\dfrac{\frac{1}{\sqrt{(t+1)!}}}{\frac{1}{\sqrt{t!}}}=\sqrt{\frac{t!}{(t+1)!}}.$$ Now apply ratio test (couple steps in between are left to you).

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