New answers tagged

2

Suppose a number $N$ has prime decomposition $N=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ We recognize that due to the properties of prime numbers, every divisor of $N$ can be written in the form $p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}$ with $0\leq \beta_i\leq \alpha_i$ for every $i$. We recognize further that every tuple of the form $(\...


0

OK, for what it's worth, the answer can be derived by casting out all factors of $5$ and multiplying the resulting numbers incrementally $\bmod 100000$, which also allows us to also divide out from the running product the same power of $2$ as the power of $5$ that we just cast out. As a way of checking any more elegant mathematical approach, therefore, the ...


2

Your corrected conjecture is true. Suppose $p$ is the smallest prime dividing $n\in\mathbb N$ and suppose $kn+ap=m!$, where $ap<kn$, for some $a,k,m\in\mathbb N$, then there are $k^{\prime},a^{\prime}\in\mathbb N$ such that $k^{\prime}n+a^{\prime}p=(m+1)!$, where $a^{\prime}p<k^{\prime}n$. Taking $k'=(m+1)k$ and $a'=a(m+1)$ works since $$k'n+...


6

For $p=3$, $q$ has to be $2$. Suppose that there exist $k,m\in\mathbb Z$ such that $$3k+2=m!$$ Since $m\gt 2$, the RHS is divisible by $3$. This is a contradiction. Added : Similarly, for $p=5$, there is no such prime $q$.


1

To be a perfect square, every prime factor has to appear an even number of times. $199$ is prime and is a factor of $199!$ and $200!$, so it appears an even number of times in the original product. If you erased $199!$ or $200!$ the product could not be square as it would have an odd number of $199$s. $197$ is prime. Which factorials does it appear in? ...


6

Divide $128$ items in half, and assign one half a $1$ bit in the first digit and the other a $0$ bit. Then divide each half in half again, and in each half assign one half a $1$ bit in the second digit and the other a $0$ bit. Continue until the halves consist of single elements. Now each element has been assigned a binary number from $0$ to $127$. The left-...


20

For fun, we do it using a combinatorial argument. Take $128$ different objects. We show that the right-hand side counts the permutations of these objects. Imagine doing the permutation as follows. First decide who will be in the first half (and therefore who will be in the second half). This can be done in $\binom{128}{64}$ ways. Now decide who among the ...


12

The question is straightforward. We have the right hand side equal to $$\frac{128!}{64!^2}\frac{64!^2}{32!^4}\frac{32!^4}{16!^8}\frac{16!^8}{8!^{16}}\frac{8!^{16}}{4!^{32}}\frac{4!^{32}}{2!^{64}}\frac{2!^{64}}{1}=128!$$


6

It is simple algebraicly. If we plug in the standard formula $\binom{n}{r} = \frac{n!}{r! (n-r)!}$, then we have \begin{align*} &\binom{128}{64} \binom{64}{32}^2 \binom{32}{16}^4 \binom{16}{8}^8 \binom{8}{4}^{16} \binom{4}{2}^{32} \binom{2}{1}^{64} \\[8pt] = {} & \left(\frac{128!}{64!^2}\right) \left(\frac{64!}{32!^2}\right)^2 \left(\frac{32!}{...


3

Why does $0!$ exist? The ultimate reason that $0!$ is allowed to exist is because mathematicians define it to exist. We simply state $0!=1$ and continue from there. There are reasons why we do this, which others have expounded upon elsewhere, but ultimately there is nothing stopping us from defining whatever we like. You only see the useful definitions ...


1

For zero is just conventional to set the empty multiplication to 1 as we set the empty sum to 0. To define the factorial function with the help of the gamma function. https://en.wikipedia.org/wiki/Gamma_function Basically this is the only function that agrees with the factorial in the natural numbers, and behaves nicely.


0

Let $[x]$ denote the largest integer not exceeding $x.$ For $n\geq 1$ we have $$\log n! =\int_1^{n+1}\log [x]\; dx<\int_1^{n+1}\log x\; dx=-n+(n+1)\log (n+1)$$ and $$\log n!=\int_1^n \log (1+[x]) \;dx\geq \int_1^n\log x \;dx=1-n+n\log n.$$ So $$1/n\leq 1+\log ( (n!^{1/n}/n)<(1+1/n)\log (n+1)-\log n=\log (1+1/n)+(1/n)\log (n+1).$$ Since $(1/n)\log (n+...


4

Michael Hardy’s computational proof is the simplest, but there is also a reasonably straightforward combinatorial argument. There are $(n+1)!-1$ permutations of the set $[n+1]=\{1,2,\ldots,n+1\}$ other than the increasing permutation $\langle 1,2,\ldots,n+1\rangle$. Let $P$ be the set of all such permutations; we’ll now calculate $|P|$ in another way. For $...


1

I think, that an "analytical continuation" is impossible for this series because the powerseries in $x$ has zero-convergence radius and thus the method of recentering the series to extend its evaluatable range step-by-step cannot be exploited here. Because you said you like the problem of divergent series - here some (amateurish, but I think: really nice) ...


0

How about ${2\choose 1}{14\choose 3}{15\choose 2}=76440$ i.e. we select one of the two girls with 3 from 14 boys and then we have to select the remaining 2 girls from 15


1

If you expand $$(3n+3)!=(3n+3)\times(3n+2)\times(3n+1)\times(3n)!$$ So $$\frac{(3n)!}{(3n+3)!}=\frac 1 {(3n+3)\times(3n+2)\times(3n+1)}$$


1

they are not same , consider $n=1$ , you have $\frac{1}{3!}=\frac{1}{6}$ and on the other hand $\frac{3!}{6!}= \frac{1}{120}$


2

The given inequality is $x!-y!-x^n \geq 0$ Observe that $x! \geq 2(x-1)!$ So if we can prove, $[(x-1)!-y!]+[(x-1)!-x^n] \geq 0$, we are done. $(1)$ $(x-1)!-y! \geq (2y-1)! -y! \geq 0$ (as $x \geq 2y$) $(2)$ $(x-1)! -x^n \geq (x-1)! -x^{(x-2)/2}$ [as $x \geq 2y, y >n \Rightarrow x \geq 2n+2 \Rightarrow n \leq \frac{x-2}{2}$] So we are left to prove $...


28

An alternative approach to Behrouz' fine one through Euler's beta function. We have: $$\begin{eqnarray*} \sum_{n\geq 0}\frac{n!}{(n+4)!}=\sum_{n\geq 0}\frac{\Gamma(n+1)}{\Gamma(n+5)}&=&\frac{1}{\Gamma(4)}\sum_{n\geq 0}B(4,n+1)\\&=&\frac{1}{6}\int_{0}^{1}\sum_{n\geq 0}x^{n}(1-x)^3\,dx\\&=&\frac{1}{6}\int_{0}^{1}(1-x)^2\,dx\\&=&\...


-1

$A=\frac{1}{4*6}+\frac{1}{10*12}+\frac{1}{18*20}+\frac{1}{28*30}+\frac{1}{40*42}+\frac{1}{54*56}...$ $2A=\frac{2}{4*6}+\frac{2}{10*12}+\frac{2}{18*20}+\frac{2}{28*30}+\frac{2}{40*42}+\frac{2}{54*56}...$ $2A=\frac{1}{4}-\frac{1}{6}+\frac{1}{10}-\frac{1}{12}+\frac{1}{18}-\frac{1}{20}+\frac{1}{28}-\frac{1}{30}+\frac{1}{40}-\frac{1}{42}+\frac{1}{54}-\frac{1}{...


6

Just a tip; For natural number $n$, $$\sum_{k=1}^n\frac{1}{k(k+1)\cdots(k+m)}=\frac{1}{m}\left(\frac{1}{1\cdot 2\cdot \cdots\cdot m} -\frac{1}{(n+1)\cdot (n+2)\cdot \cdots\cdot (n+m)}\right) $$ Now, the given infinite sum is equal to $$\sum_{k=1}^n \frac{(k-1)!}{(k+3)!}=\sum_{k=1}^n \frac{1}{k(k+1)(k+2)(k+3)}$$ by plugging in $m=3$ to the tip, we would ...


27

Write $$\begin{align} {\frac{n!}{(n+4)!}}&=\frac{1}{(n+1)(n+2)(n+3)(n+4)}=\\&=\frac{A}{(n+1)(n+2)(n+3)}-\frac{B}{(n+2)(n+3)(n+4)}\end{align}$$ which gives $(n+4)A-B(n+1)=1 \implies A=B=1/3$ Then we get a telescoping series: $$\begin{align}\sum\limits_{n=0}^{\infty }{\frac{n!}{(n+4)!}}=\frac{1}{3}\sum\limits_{n=0}^{\infty }\left({\frac{1}{(n+1)(n+2)...


1

I'd suggest some combinatorial argument. Consider a set of $n$ balls, where among them, there are $m$ kinds of balls. And there are $p_i$ number of balls of type $i$, i.e. $p_1$ number of balls of $1$-st kind, $p_2$ number of balls of $2$-nd kind, and so on, like $p_m$ number of balls of $m$-th kind. Moreover, each kind of balls are identical, where ...


1

Quite an overkill, but since $n!=\Gamma(n+1)$, your limit is a consequence of Gautschi's inequality.


1

$$\frac{n!}{(n-k)!} = n(n-1)\cdots(n-(k+1))$$ is a polynomial in $n$ of degree $k$. So you have the limit towards infinity of a rational function whose numerator and denominator have the same degree. Thus the limit is the ratio of the leading coefficients. $$\lim\limits_{n\to\infty} \frac{\left(\frac{n!}{(n-k)!}\right)}{n^k} = 1$$


0

Hint: $$\lim_{n\to\infty}\dfrac n{n-k}=?$$


6

(There's no need for the superscripts on elementary symmetric polynomials $e_i$.) As Vieta's formula states, $e_i(x_1,\cdots,x_n)$ will be the $i$th coefficient of $(T+x_1)\cdots(T+x_n)$. According to Euler's theorem, $x^{p-1}\equiv 1$ mod $p$ for all $x$, so the polynomial $T^{p-1}-1$ has every integer residue $1,\cdots,p-1$ as a zero. Since the integers ...


1

Assuming $G>0$, by using the Taylor series of the exponential function and the identity $m!=\int_{0}^{+\infty}x^m e^{-x}\,dx$ we have: $$\begin{eqnarray*}\sum_{k\geq 0}\frac{(G+k)!}{2^k(k+1)!}&=&\sum_{k\geq 0}\frac{1}{2^k(k+1)!}\int_{0}^{+\infty}x^{G+k}e^{-x}\,dx\\[0.2cm]&=&\int_{0}^{+\infty}x^G e^{-x}\sum_{k\geq 0}\frac{x^k}{2^k(k+1)!}...


3

You can simply write it like this also this solution doesn't need induction: $=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n+1-1}{(n+1)!}$ $=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{n!}-\frac{1}{(n+1)!}$ $=1-\frac{1}{(n+1)!}$ $$solved!$$


1

We have $$S=\sum_{k\geq0}\frac{\left(G+k\right)!}{\left(k+1\right)!}\left(\frac{1}{2}\right)^{k}=\left(G-1\right)!\sum_{k\geq0}\dbinom{G+k}{G-1}\left(\frac{1}{2}\right)^{k} $$ and since holds $$\dbinom{G+k}{G-1}=\sum_{m=0}^{G-1}\dbinom{k+m}{m} $$ we have, exchanging the sum with the series $$S=\left(G-1\right)!\sum_{m=0}^{G-1}\sum_{k\geq0}\dbinom{k+m}{m}\...


2

Generating functions to the rescue! \begin{align*} \sum_{k=0}^\infty x^k &= \frac1{1-x} \\ \sum_{k=0}^\infty x^{G+k} &= \frac{x^G}{1-x} = \frac1{1-x} - 1 - x - \cdots - x^{G-1} \\ \frac{d^{G-1}}{dx^{G-1}} \sum_{k=0}^\infty x^{G+k} &= \frac{d^{G-1}}{dx^{G-1}} \bigg( \frac1{1-x} - 1 - x - \cdots - x^{G-1} \bigg) \\ \sum_{k=0}^\infty (G+k)_{G-1} x^{...


6

I thought it might be instructive to present an approach that is not a proof by induction. To that end, note that we can write $$\bbox[5px,border:2px solid #C0A000]{\frac{k}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}}$$ Therefore, we have a telescoping sum such that $$\begin{align} \sum_{k=1}^n \frac{k}{(k+1)!}&=\sum_{k=1}^n\left(\frac{1}{k!}-\frac{1}...


3

You have your base case. Explicitly state your inductive hypothesis. Suppose: $\sum_\limits{n=1}^k\frac{n}{(n+1)!} = 1-\frac{1}{(k+1)!}$ We will show that: $\sum_\limits{n=1}^{k+1}\frac{n}{(n+1)!} = 1-\frac{1}{(k+2)!}$ $\sum_\limits{n=1}^k\frac{n}{(n+1)!}+\frac{(k+1)}{(k+2)!}$ $1-\frac{1}{(k+1)!}+\frac{(k+1)}{(k+2)!}$(By the inductive hypothesis.) $1-\...


4

The equality you want to prove is $$ \underbrace{\frac{1}{2!}+\dots+\frac{k}{(k+1)!}}_{*}+ \frac{k+1}{(k+2)!}=1-\frac{1}{(k+2)!} $$ The term marked $*$ is equal, by the induction hypothesis, to $$ 1-\frac{1}{(k+1)!} $$ and so you need to manipulate $$ 1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!} $$ Hint: $$ 1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!} = 1-\frac{k+2}{(k+2)!...


0

Assume that $x>a>0$. Then: $$\frac{x!}{a^x}=\frac{a!\Pi^x_{i=a+1}i}{a^x}>a!\frac{(a+1)^{x-a}}{a^x}=\frac{a!}{(a+1)^a}\frac{(a+1)^x}{a^x}=\frac{a!}{(a+1)^a}(1+\frac{1}{a})^x\to_{x\to\infty}\infty$$


1

In particular, for $\displaystyle k=\frac{1}{2}$, \begin{align*} \Gamma \left( n-\frac{1}{2} \right) &= \frac{(2n-3)!!}{2^{n-1}} \sqrt{\pi} \\ \frac{\Gamma \left( n \right)}{\Gamma \left( n-\frac{1}{2} \right)} &= \frac{(n-1)!}{(2n-3)!!} \frac{2^{n-1}}{\sqrt{\pi}} \\ &= \frac{(n-1)!(2n-2)!!}{(2n-2)!!(2n-3)!!} \frac{2^{n-1}}{\sqrt{\pi}} \\...


17

Look at the prime factors of $n!$. If the square root of $n!$ was an integer, then $n!$ would be the square of an integer, and in the square of an integer, all prime factors occur an even number of times. For example, if you take $100!$, which ends with $\cdots 95\times 96\times 97\times 98\times 99\times 100$, you see the prime number $97$. That prime ...


51

For any $n \gt 1$ there will be some prime in the range $(n/2,n]$ which will only occur once in the factorization of $n!$ by Bertrand's Postulate. This will ensure that $\sqrt{n!}$ is not an integer.


5

I will repeat basically the same approach as in this answer: Limit $\lim\limits_{n\to \infty} \sqrt [n]{\frac{(3n)!}{n!(2n+1)!}} $ It is also the same approach as suggested in Paramanand Singh's comment. (I see that the OP asks specifically about a proof using Riemann's integral, but this seems interesting enough to be mentioned, too.) We will use this fact ...


-5

We define factorial using recursive piecewise functions. $$x! = x*(x-1)!*I[x > 0] + 1*I[x=0]$$ You might notice that only the term where x is 0 lacks a recursive use of factorial. This term exists solely to end the recursion. Otherwise, it would be an infinitely recursive function. Theres no proof, and i think factorial of 0 would make sense as 0. ...


-1

$n!$ is defined as the number of way n distinct object can be arranged. For example, $1!$ is described to arrange $1$ object,i.e $1$ way. $A$ $2!$ is described to arrange $2$ object,i.e $2$ way. $AB,BA$ $3!$ is described to arrange $3$ object,i.e $6$ way. $ABC,ACB,BAC,BCA,CAB,CBA$ Similary, $0!$ describe the way to arrange $0$ object.$0$ object means ...


4

Let's try splitting up the series (since if it converges to a finite number, it will be absolutely convergent, since all terms are positive). $$\sum_{k=1}^{\infty} \frac{1}{(k+1)(k-1)!}(1 - \frac{2}{k}) = \sum_{k=1}^{\infty} \frac{1}{(k+1)(k-1)!} - 2 \sum_{k=1}^{\infty} \frac{1}{(k+1)(k)(k-1)!} \\ = \sum_{k=1}^{\infty} \frac{k}{(k+1)(k)(k-1)!} - 2 \sum_{k=1}...


6

Hint: Note that $1-\frac{2}{k}=\frac{(k+1)-3}{k}$. So our sum is $$\sum_1^\infty \frac{1}{k!} -3\sum_1^\infty \frac{1}{(k+1)!}.$$ Each sum is a fairly close relative of $e$.


2

Since $\frac{\Gamma(x+n)}{\Gamma(x)}=(x)_n$, the answer is simply given by the generating function for Stirling numbers of the first kind.


14

Applying Stirling's formula, $n! \displaystyle\operatorname*{\sim}_{n\to\infty} \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$, I got the result $\frac{4}{e}$.


23

Hint: take the logarithm, and try to make something like $\frac{1}{n}\sum_{k=1}^n f\left(\frac{k}{n}\right)$ appear for $f\colon x\in[0,1] \mapsto \ln (1+x)$. You will find the limit $\ell$ of the logarithm of your quantity, and then by continuity of $\exp$ your answer will be $e^\ell$. Details. Taking the logarithm, $$\begin{align} \ln \sqrt[n]{\frac{(...


4

I take it that $k$ is a positive integer. Then the question is equivalent to $e^k>\frac{k^k}{k!}$ But $e^k=\displaystyle \sum_{n=0}^\infty \frac{k^n}{n!}$ and one of the summands (all of them are positive) is itself $\frac{k^k}{k!}$


2

Work by induction. For $k=1$ it is true as $e >1$ Assume it is true for $k$. Then $(k+1)! = (k+1) k! > k k!> k \frac{k^k}{e^k} = e \frac{k^{k+1}}{e^{k+1}} > \frac{k^{k+1}}{e^{k+1}}$ because $e>1$.



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