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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.


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)...


27

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\\&=&\...


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{(...


18

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 ...


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 ...


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 $\frac{4}{e}$.


12

Here's another non-induction and non-combinatorial proof. $$\begin{align} \sum_{j=0}^{s}(n-s+j)!\binom{s}{j}(s-j)! &=s! \sum_{j=0}^{s} \frac{(n-s+j)!}{j!} \cdot \color{lightgrey}{\frac{(n-s)!}{(n-s)!}}\\ &=s!(n-s)!\sum_{j=0}^s\binom {n-s+j}{n-s}\\ &=s!(n-s)!\binom {n+1}{n-s+1} &&\tiny\text{using }\sum_{r=a}^b\binom {m+r}{m+a}=\binom {m+...


9

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


8

I am getting: $$\begin{eqnarray*}\sum_{j=0}^{s}\binom{s}{j}(n-s+j)!(s-j)!&=&\iint_{(0,+\infty)^2}\sum_{j=0}^{s}\binom{s}{j}x^{s-j} y^{n-s+j} e^{-(x+y)}\,dx\,dy\\&=&\iint_{(0,+\infty)^2} x^s y^n e^{-(x+y)}\left(\frac{1}{x}+\frac{1}{y}\right)^s\,dx\,dy\\&=&\iint_{(0,+\infty)^2} y^{n-s} e^{-(x+y)}(x+y)^s\,dx\,dy\\&=&\int_{0}^{+\...


7

I use the notation $x^{\underline k}=x(x-1)\ldots(x-k+1)$ for the falling factorial. First note that $$\sum_{j=0}^s(n-s+j)!\binom{s}j(s-j)!=\sum_{j=0}^s(n-s+j)!s^{\underline{s-j}}=\sum_{j=0}^ss^{\underline j}(n-j)!\;.\tag{1}$$ Let $A=\{0,1,\ldots,n\}$; then $s^{\underline j}(n-j)!$ is the number of permutations $a_0a_1\ldots a_n$ of $A$ such that $a_j=s$, ...


6

$34! = 295232799cd96041408476186096435ab000000$ $\left \lfloor \dfrac{34}{5} \right \rfloor = 6$ $\left \lfloor \dfrac{6}{5} \right \rfloor = 1$ So there are $6+1 = 7$ zeros at the end of $34!$. Hence $$\color{red}{b = 0}$$ THEOREM: Compute the following $N = 5q_1 + R_1$ $q_1 = 5q_2 + R_2$ $q_2 = 5q_3 + R_3$ ... $q_{n-1} = 5q_n + R_n$ where $0 \le ...


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 ...


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}...


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 $e$.


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 ...


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 $q$.


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

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!}{...


5

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-...


4

Let me first summarize the observations you already made yourself, using the following notations: $S_n=1+1+2!+\ldots + n!$, $d_n=(S_n,(n+1)!)$ Then $2|d_n$, $2^2\nmid d_n$; the second property follows $2^2\nmid S_3$, but $2^2|(n+1)!$ for $n\ge 3$ and $S_{n+1}=S_n+(n+1)!$ so $2^2\nmid S_n$ for all $n$ by induction. Also, $d_n=2$ initially; a new prime $p$ ...


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 $\frac{k^k}{k!}$


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}...


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)!...


4

$$ n!\cdot n \quad = \quad n!\Big((n+1) - 1\Big) \quad = \quad \big(n!(n+1) \big) - \big( n!\cdot1 \big) \quad = \quad (n+1)! - n!. $$ Therefore \begin{align} & 1!1+2!2 + 3!3 + 4!4 + 5!5 \\[10pt] = {} & \Big(2!-1!\Big) + \big( 3!-2!\Big) + \Big(4!-3!\Big) + \Big(5!-4!\Big) + \big(6!-5!\Big) \end{align} Now notice that all terms cancel except the ...


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 $...


3

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


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-\...


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 ...


2

This is a partial solution... Use all the divisibility rules. But first, you can notice that $32!=2^31\times5^7\times\cdots$, so there are $7$ zeros at the end of the number. So $b=0$, and $a\ne 0$. $34!$ is divisible by $9$, so: $$4+a+c+d=0\pmod 9.$$ It's divisible by $7$, so: $$000-000+5a0-643+609-618+847-140+604-cd9+799+327-952+2=0\pmod 7,$$ so ...



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