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10

You can use the identity given by the Euler Beta function $$\int_{0}^{1}x^{a-1} (1-x)^{b-1} \,dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ to state: $$S=\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k!}\Gamma(k/2)^2=\sum_{k=1}^{+\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}\left(x(1-x)\right)^{k/2-1}\,dx$$ and by switching the series and the integral: $$S = ... 9 The number of zeros at the and of n! is just the number of factors of 5 in n! (since the number of factors of 2 in n! is always larger than that). So, we need all n such that the number 1 up to n have a total of 30 factor of 5. A first guess would be 30\cdot 5=150, but then we forget that the multiples of 5^2=25 have (at least) two ... 6 Note that$$ \begin{align} \frac{x}{1-x} &=\sum_{n=1}^\infty x^n\\ &=\sum_{n=1}^\infty\sum_{k=1}^n\frac{a_k}{(n-k+1)!}x^n\\ &=\sum_{k=1}^\infty\sum_{n=k}^\infty\frac{a_k}{(n-k+1)!}x^n\\ &=\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{a_k}{(n+1)!}x^{n+k}\\ &=\frac{e^x-1}{x}\sum_{k=1}^\infty a_kx^k\tag{1} \end{align} $$Therefore,$$ ...

5

Your conjecture is true. We have the following celebrated result in math olympiads: Theorem. (Lifting The Exponent) If $p$ is an odd prime, $n\in\mathbb N$, $a,b\in\mathbb Z$ such that $p\mid a-b$ but $p\nmid a,b$, then $\nu_p(a^n-b^n)=\nu_p(n)+\nu_p(a-b)$, where $\nu_p(x)$ denotes the number of factors $p$ in the prime factorisation of the integer ...

4

Note that $$(n!)^2 = \prod_{i=1}^n i(n+1-i) \ge \prod_{i=1}^n n = n^n$$ Hence $n! \ge n^{n/2}$. This gives $$(2n)! \ge (2n)^n = 2^n n^n$$ For $n \ge 2$ we have $2^n \ge n^2$, hence $(2n)!\ge n^{n+2}$.

4

Put $\displaystyle f(x)=\sum_{k\geq 1} a_k x^{k-1}$, $\displaystyle g(x)=\sum_{l\geq 0} \frac{x^l}{(l+1)!}=\frac{\exp(x)-1}{x}$, and $\displaystyle h(x)=f(x)g(x)=\sum_{n\geq 0}c_n x^n$. The coefficient $c_n$ is equal to $1$ if $n=0$, and for $n\geq 1$: $$c_n=\sum_{k-1+l=n, k\geq 1, l\geq 0}\frac{a_k}{(l+1)!}=\sum_{k=1}^{n}\frac{a_k}{(n+1-k)!}=1$$ Hence: ...

4

You just need to do it once and from that, get a recursion relation: \begin{align*} I_n &= \int_{x=0}^\infty x^n e^{-x} \, dx \\ &= \lim_{t \to \infty} \Bigl[ -x^n e^{-x} \Bigr]_{x=0}^t + n \int_{x=0}^\infty x^{n-1} e^{-x} \, dx \\ &= \lim_{t \to \infty} - t^n e^{-t} + n I_{n-1} \\ &= n I_{n-1} . \end{align*} Then it is trivial to ...

3

The asymptotic expansion for $\Gamma(z)$ is given by $$\Gamma(z) \sim \sqrt{\frac{2\pi}{z}}\left(\frac{z}{e}\right)^z\left(1 + \frac{1}{12z} + \cdots\right)$$ Therefore we have \begin{align}\Gamma\left(n+\frac{1}{2}\right) &\sim \sqrt{\frac{2\pi}{\left(n+\frac{1}{2}\right)}}\left(\frac{n+\frac{1}{2}}{e}\right)^{n+\frac{1}{2}} \\&= ... 3n!=1 \cdot 2 \cdot 3 \cdots nn!+3=1 \cdot 2 \cdot 3 \cdots n+3=3(1 \cdot 2 \cdot 4 \cdots n+1)$$We see that 3 \mid n!+3 \Rightarrow n!+3=3k, k \in \mathbb{Z} n \geq 3 \Rightarrow n! \geq 6 \Rightarrow n!+3 \geq 9, \text{ that implies that } k \geq 3. So, n!+3 is composite. 2 Hint Prove that 3 divides n!+3. 2 (This is more of a comment that an answer but, ...) Consider the even and odd k in separate sums: Note: You probably do not want \Gamma(0) in the sum, so I'll start at k=1.$$\begin{array}\\ S &=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\left[\Gamma\left(\frac{k}{2}\right)\right]^2\\ ...

2

There are three choices for the first digit. Once this choice has been made, there are four digits left. These can be arranged in $4!$ ways. Then $3\times 4!$ gives the correct answer. Added Later: From what I can tell, you made two mistakes. First of all, after the choice of the first digit, you said there were three numbers left when in fact there are ...

2

We have four groups of three arranged around the circle. First choose the four seats of the managing directors. This can be done in $3$ ways. Next, distribute the four companies around the four quarters of the circle (around the four chosen seats). That can be done in $4!=24$ ways. Next, for each company, choose who sits to the left an who sits to the right ...

1

It is known as factorial and denoted as $n!$. The case $10!$ can be reduced: \begin{eqnarray} 1 \cdot 2 \cdot 3 \cdots \cdot 8 \cdot 9 \cdot 10 &=& 10 \Big(5-4\Big) \Big(5-3\Big) \cdots \Big(5+4\Big)\\ &=& 50 \Big(25-16\Big) \Big(25-9\Big) \Big(25-4\Big) \Big(25-1\Big)\\ &=& 50 \Big(15-6\Big) \Big(15+6\Big) \Big(20-4\Big) ...

1

There are $24!$ permutations of the letters b through y. For each such permutation, if a is to the left of z, it can appear in any of 15 positions (from before the first letter to before the fifteenth letter), and z appears 10 positions later. Similarly, if z is first, it can appear in any of 15 positions and a appears 10 positions later. So $30\cdot 24!$.

1

Several problems: Using $P(26,10)$ includes arrangements where a and z are not 10 apart. There are 24 letters excluding a and z. Once the letters between a and z are chosen, there are $26-(10+2)=14 \neq 16$ remaining letters. The number of letters to the left of a. It is possible to have anywhere between $0$ and $26-(10+2)$ letters to the left of a. a ...

1

If you take the natural log we get $$\ln(a_n)=(n+1)\ln(n+1)+\ln(n!)-( (n+1)!-n! )\ln(2)$$ $$= (n+1)\ln(n+1)+\sum_{1}^{n}\ln(k)-n\,n!\ln(2)$$ $$= (n+1)\ln(n+1)+n\ln(n)-n+1-n\,n!\ln(2) \sim_{ \infty} 2n\ln(n)-n-nn!\,\ln(2)$$ $$= 2n\ln(n)-(n+1)!\,\ln(2) \to -\infty$$ $$\implies a_n \rightarrow_{n\to\infty} 0$$ Note: $$\sum_{1}^{n}\ln(k) ... 1 The correct formula for the number of trailing zero digits in n! is$$ \frac{n-\sigma_5(n)}{4} $$where \sigma_5(n) is the sum of the base-5 digits of n. So the formula given to you is only correct if the sum of the base-5 digits of n is 4. Since$$ \begin{array}{l} 100_\text{ten}=400_\text{five}\\ 200_\text{ten}=1300_\text{five}\\ ...

1

The factorial function is only defined on the positive integers, so those don't make sense. However, there is a generalization of the factorial called the Gamma function which you might want to check out.

1

This is a very amazing problem. As I said earlier, I did not find any closed form for the integral and then I only performed numerical integrations. What is surprizing if that $y(a)=\displaystyle\int_{0}^{a} \Gamma(1+x)~ dx$ looks very much to $x(a)=\Gamma(1+a)$ (plot the two curves as a function of $a$). Taking into account Lucian's comment, I performed a ...

1

This is more a comment that an answer As Jack D'Aurizio commented, I suppose that the summation starts at $k=1$ and not $k=0$ as written in the post. Using a CAS, the result obtained is $$\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\left[\Gamma\left(\frac{k}{2}\right)\right]^2=\frac{5 \pi ^2}{18}$$ which matches the value you obtained using Wolfram Alpha. ...

1

Here is a closed form $$\mathcal{I}=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right)x^k= 2 \arcsin \left( x/2 \right) \left(\pi - \arcsin \left( x/2\right) \right) .$$ Now just plug in $x=1$ and the result follows.

1

Just that you can check much higher; you know how to find $b_n.$ Then calculate $$3^{n!} - 1 \pmod {2^{b_n + 2}},$$ $$3^{n!} - 1 \pmod {5^{b_n + 2}}.$$ I don't see much reason that (1) should hold forever, but up to 100 would be impressive. There are topics sort of like this where there is agreement for an initial set but later disagreement. The ...

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