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9

$$\frac{(m+n)!}{m!n!}=\binom{m+n}{n}>1$$


8

Note that $$n!\,e=\sum_{k=0}^\infty\frac{n!}{k!}=\sum_{k=0}^n\frac{n!}{k!}+\sum_{k=n+1}^\infty\frac{n!}{k!}$$ The first sum on the RHS is always an integer since $n\geq k$. The second sum satisfies $$\begin{align} \sum_{k=n+1}^\infty\frac{n!}{k!} &=\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots\\ ...


8

Hint: for combinatorial reasons (in how many ways can you choose $n$ objects from $n$ pairs of objects?), $$\binom{2n}n\ge2^n\ ,$$ so $$\frac{(n!)^2}{(2n)!}=\frac1{\binom{2n}n}\le\frac1{2^n}\ .$$


7

$$\frac{10^n}{n!} = \frac{10\cdot 10 \cdot 10 \cdot 10 \cdots 10}{1 \cdot 2 \cdot3 \cdots n} =\frac{10^{11}}{11!}\frac{10}{12}\frac{10}{13}\cdots\frac{10}n < \frac{10^{11}}{11!}\left(\frac{5}{6}\right)^{n-11} \to 0$$ as $n \to \infty.$


7

Well, for starters you can cancel the two $(n+1)!$s from the top and the bottom of the fraction. Also note that $(n-1)! = (n-1)((n-2)!)$ and then you can cancel an $(n-2)!$ from the top and the bottom.


6

As all terms are positive, we have $$0 \leq \frac{(n!)^2}{(2n)!} = \frac{n!}{2n \cdot \dots \cdot (n+1)} = \prod_{k=1}^n \frac{k}{k+n} \leq \prod_{k=1}^n \frac{1}{2} = \left(\frac{1}{2}\right)^n$$ So then as $$\lim\limits_{n\rightarrow\infty} \left(\frac{1}{2}\right)^n = 0$$ It follows that $$\lim\limits_{n\rightarrow\infty} \frac{(n!)^2}{(2n)!} = 0$$ ...


6

By Stirling's Approximation, we have $$x! \approx \sqrt{2 \pi x} \left(\frac{x}{e}\right)^x,$$ and so for large $n$ $$(x!)^{1 / x} \approx (2 \pi)^{1 / 2x} x^{1 / x} \frac{x}{e}.$$ Thus, $$[(x + 1)!]^{1 / (x + 1)} - (x!)^{1 / x} \approx (2 \pi)^{1 / 2(x + 1)} (x + 1)^{1 / (x + 1)} \frac{(x + 1)}{e} - (2 \pi)^{1 / 2x} x^{1 / x} \frac{x}{e}.$$ Now, $(2 \pi)^{1 ...


5

Now we see that $(3!, 11)=(6,11)=1$. Hence by Fermat's theorem we have $(3!)^{10}\equiv 1[11]$ and hence $(3!)^{10m}\equiv 1[11]$. Moreover $10$ divides $5!$ so that $(((3!)^{5!})^{7!})^{...}\equiv 1[11]$. Required reminder is 1.


5

If you really want to prove this identity using bloody/gory algebraic manipulations, then note the following: \begin{align} \binom{m}{n}+\binom{m}{n-1} &= \frac{m!}{n!(m-n)!}+\frac{m!}{(n-1)!(m-n+1)!}\\[1em] &= \frac{m!}{n!(m-n)!}+\frac{n\cdot m!}{n!(m-n)!(m-n+1)}\\[1em] &= \frac{m!(m-n+1)+n\cdot m!}{n!(m-n)!(m-n+1)}\\[1em] &= ...


5

Lazy people like me should do it as follows. Define $R(m,n)$ as: $$R(m,n) = \frac{\binom{m}{n}+\binom{m}{n-1}}{\binom{m+1}{n}}$$ It should be clear that $R(m,n)$ is some rational function and we want to prove that this is actually identical to 1. Now, because it is manifestly a rational function, the algebra is guaranteed to simplify a lot making this ...


4

$$(3!,11)=1$$ By Fermat's Little Theorem, $$3^{11-1}\equiv1\pmod{11}$$ and $10|5!$


4

They are using the trapezoid rule over $n-1$ intervals: $(1,2), (2,3) \dots (n-1,n)$ The trapezoid rule for the first interval is $\int_1^2f(x)\;dx \approx \frac 12(f(1)+f(2))$ When you add these up over all the intervals, each internal point gets a coefficient $1$, with $\frac 12$ coming from the interval below and $\frac 12$ from the interval above but ...


4

Hint: for any $\;x\in\Bbb R\;$ : $$\sum_{n=0}^\infty\frac{x^n}{n!}=e^x$$


4

Recalling the Stirling's approximation $$n!\sim\sqrt{2\pi}n^{n+1/2}e^{-n} $$ we have $$\sqrt{2\pi}\lim_{n\rightarrow\infty}\frac{n^{2n+1}e^{-2n}}{\left(2n\right)^{2n+1/2}e^{-2n}}=0. $$


4

Let's rewrite this as $\;\displaystyle S=\sum _{j=0}^{n} \frac{1}{j!}-2\;$ then prove that $$\sum _{j=0}^{n} \frac{1}{j!}=\frac{\lfloor e\; n!\rfloor}{n!}$$


4

The series $\sum_{n=0}^\infty \frac{10^n}{n!}$ is convergent and the sum is $e^{10}$. The convergence can be checked by ratio test $$\lim_{n \to \infty} \frac{10^{n+1}}{(n+1)!}\frac{n!}{10^n}=\lim_{n \to \infty}\frac{10}{n+1}=0<1$$ By the necerssery condition for the convergence of an infinite series the common term must go to 0.


3

Stirlings approximation of the factorial: $$ \sqrt{2\,\pi\,n}\left(\frac{n}{e}\right)^n $$ 5 and 3 are constant factors so put them before the limit. Now we have to concern ourselves with: $$ \frac{5}{3}\lim_{n\rightarrow\infty}\dfrac{n^n}{\sqrt{2\,\pi\,n}\left(\frac{n}{e}\right)^n+3^n} = ...


3

This looks very similar to Pochhammer symbol used to represent the falling factorial $$(x)_n=x(x-1)(x-2)\cdots(x-n+1)$$ and when $x$ is a non-negative integer, $(x)_n$ gives the number of $n$ permutations of an $x$ element set, or equivalently the number of injective functions from a set of size $n$ to a set of size $x$ (this is a quote from the ...


3

From DeMoivre's formula, we have $$x! \sim C \sqrt{x}\left(\dfrac{x}e\right)^x + \mathcal{O}(1/x)$$ This means we have \begin{align} ((x+1)!)^{1/(x+1)} - (x!)^{1/x} & \sim C^{1/(x+1)} \sqrt{x}^{1/(x+1)} \left(\dfrac{x+1}e \right) - C^{1/x} \sqrt{x}^{1/x} \left(\dfrac{x}e\right) + \mathcal{O}(1/x)\\ & \sim \dfrac{x+1-x}e = \dfrac1e \end{align}


3

Hint: $$(k+1)!> (k+1)^2$$ can be rewritten as $$(k+1)\cdot (k!) > (k+1)\cdot(k+1)$$


3

If $x$ is an integer, we can use Sterling's formula $$x! \approx. \sqrt{2\pi x} \left(\frac{x}{e}\right)^x$$ Thus, $$((x+1)!)^{1/(x+1)}\approx. (\sqrt{2\pi (x+1)})^{1/(x+1)}\left(\frac{x+1}{e}\right)$$ and $$(x!)^{1/x}\approx. (\sqrt{2\pi x})^{1/x}\left(\frac{x}{e}\right)$$ The term $\sqrt{2\pi x}^{1/x}$ and $\sqrt{2\pi (x+1)}^{1/(x+1)}$ can easily be ...


3

De Polignac's formula tells us that $36!$ there are 34 powers of $2$ and 8 powers of $5$, so when $36!$ is factored, there are $8$ zeros because there is a $10^8$ in its factorisation. Since each power of $36!$ adds 8 zeroes, $36!^{36!}$ has $8\cdot36!$ zeros at the end.


3

How many ways can you pick $k$ balls from a set of $n$ different red balls and $m$ green different balls? Answer $$(n+m)^{\underline k}$$ But you can count them another way. First suppose that the $k$ balls are red, then $k-1$ are red and $1$ is green, etc. This gives $$\sum_{j=0}^k\binom kj n^{\underline j} m^{\underline {k-j}}$$


3

Oh, lol! $\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n-2)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n-1)! + (n-2)!}{3!(n-2)!} = \frac{(n-2)!((n-1) + 1)}{3!(n-2)!}=\frac{n}{3!} = \frac{n}{6}$ Correct? :) How come Wolfram Alpha gives two different results?


3

The "triple factorial" is defined for non-negative integers $n$ as $$ n!!! = n (n-3)(n-6) \cdots (n \bmod 3) \, , $$ compare "Multifactorials" in Wikipedia. For $n \ge 0$ and $j > 0$ $$ n!!! = \frac{(n+3j)!!!}{(n+3)(n+6)\cdots (n+3j)} \tag{1} $$ holds. This relation can be used to define the function for negative integers as well, as long as they are not ...


3

There is a very simple combinatorics proof for that. Assume that we have $m+1$ objects called $a_0, \dots, a_m$ and we want to select $n$ objects out of them. There are $\binom {m+1}{n}$ ways to do this. This selection can be done in two ways. Either we include the first element in our selection or not. If we include the first element we are left with ...


3

You want to prove that $2^n (n!)^2 \le (2n)!$, but this is equivalent to: $$2^n \le {2n \choose n} = \frac{(2n)!}{n! \cdot n!}$$ Suppose you have $n$ pairs, each consisting of one woman and one man. Right side counts the number of ways to choose $n$ people from that group. Left side counts the number of ways to choose one person from each pair. After ...


2

$2^n(n!)^2\le (2n)!$. Now divide by $n!$ and get: $$\iff \underbrace{2n(2(n-1))(2(n-2))\cdots(2(n-(n-1)))}_{n\text{ terms}}\le \underbrace{(2n)(2n-1)\cdots(2n-(n-1))}_{n\text{ terms}}$$ This is true because $2(n-k)\le 2n-k\iff k\ge 0$.


2

There are multiple ways. One can even use induction to prove it. Here's the easiest way: $$2^n(n!)^2=(2^n\cdot n!)\cdot n!=(2n)!!\cdot n!\le (2n)!!\cdot (2n-1)!!=(2n)!$$ $$\therefore\quad 2^n(n!)^2\le (2n)!$$ Here, $n!!$ is the double factorial notation. This is actually because $n!=1\cdot 2\cdot 3\cdots n\le 1\cdot 3\cdot 5\cdots (2n-1)=(2n-1)!!$


2

The LCM of $n!(m-n)!$ and $(n-1)!(m-n+1)!$ is $n!(m-n+1)!$. This must be the case because $n!=(n-1)!\cdot n$ and $(m-n+1)!=(m-n)!\cdot (m-n+1)$ . With this in mind, you should be able to easily convert the two fractions into a common denominator. Namely, $$\frac{m!\cdot(m-n+1)}{n!(m-n+1)!}+\frac{m!\cdot n}{n!(m-n+1)!}$$ It's really not as complicated as it ...



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