# Tag Info

10

Put $$a_n = \frac{n!^{1/n}}{n}$$ then $$\log a_n = \frac{1}{n}\left(\log n! - n\log n\right)= \frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}\right)$$ where we have used $\log n! = \log 1 + \log 2 + \ldots + \log n$. The sum above is a Riemann sum for the integral $\int_0^1\log x dx$ so $$\lim_{n\to\infty} \log a_n = \int_0^1\log x dx = [x\log x - x]_0^1 = ... 8 You have all the right pieces, but it's pretty sloppy. Here's a cleaned up version. We will prove by induction on n \in \mathbb N that:$$ 1! + 2! + \cdots + n! < (n + 1)! \tag{$\star$} $$Base Case: Notice that (\star) holds for n = 1, since:$$ 1! = 1 < 2 = (1 + 1)! $$Inductive Step: Assume that (\star) holds for n = k \geq 1. It ... 5 First, by dividing both sides by (n!)^2 and organizing the factorials into binomial coefficients we obtain the suggestive form$$\binom{2n}{n}\overset{?}{=}\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{k}\binom{n-k}{k}2^{n-2k}$$We verify this expression by a double counting argument. Consider a 2\times n array consisting of equal numbers of zeros and ... 5 You have a total of xy balls, namely y balls of each of x colors. In how many ways can you arrange these balls in a line, when two arrangements that cannot be distinguished colorwise are considered equal? 4 Proceed by induction on m. Clearly, if m=1 you have \frac{k!}{k!} = 1 is an integer. Now, for the inductive step consider$$\frac{((m+1)k)!}{(k!)^{m+1}} = \frac{(mk)!}{(k!)^m} \frac{(mk+1)(mk+2) \cdots (mk+k)}{k!}$$By hypothesis \frac{(mk)!}{(k!)^m} is an integer, so if you prove that$$\frac{(mk+1)(mk+2) \cdots (mk+k)}{k!} $$is an integer you are ... 4 Another possible approach is to use the discrete Fourier transform. Let \omega=\exp\frac{2\pi i}{3}. Then:$$f(n)=\frac{1}{3}\left(1+\omega^n+\omega^{2n}\right)=\mathbb{1}_{n\equiv 0\!\pmod{3}}(n),$$hence: ... 4 Hint: \left(n+1\right)!=n!\left(n+1\right)>n^{\frac{n}{2}}\left(n+1\right). So it is enough to prove that:$$n^{\frac{n}{2}}\left(n+1\right)\geq\left(n+1\right)^{\frac{n+1}{2}}$$or equivalently:$$n+1\geq\left(1+\frac{1}{n}\right)^{n}$$This for n>2. 4 Hint: n, (n+1), (n+2), (n+3), (n+4), (n+5) are all 6 consecutive numbers so one of them is a multiple of 6, and 3 of them are a multiple of 2; and 2 of them are a multiple of 3... 4 Hint:$$ \sum_{n=k}^m\binom{n}{k}=\binom{m+1}{k+1} $$A generalization is discussed in this answer. The equation above is equation (1) with m=0. Telescoping sum To turn the sum in the question into a "telescoping sum", we can use the recurrence for Pascal's Triangle:$$ \binom{n+1}{k+1}=\binom{n}{k}+\binom{n}{k+1} $$Using this recurrence, we get$$ ...

4

Let $$p_n=\prod_{r=1}^k (n+r)=\overbrace{(n+1)(n+2)\cdots(n+k)}^{k \ \text{terms}}$$ which is the product of $k$ consecutive integers. Consider the difference of two consecutive terms, where each term is the product of $k+1$ consecutive integers, i.e. \begin{align} &\prod_{r=1}^{k+1}(n+r)-\prod_{r=0}^k(n+r)\\ ... 4 Stirling's Approximation n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n $$Which means that n! is asymptotically equivalent to \sqrt{2\pi n}\left(\frac{n}{e}\right)^n as n approaches \infty. If your familiar with asymptotic formulas, then you'd also know that this implies that$$ \lim_{n\to\infty} \frac{n!}{\sqrt{2\pi ...

3

Also a neat way to solve it without induction $$\sum_{r=1}^nr\cdot r!=\sum_{r=1}^{n}(r+1-1)\cdot r!=\sum_{r=1}^n (r+1)!-r!=(n+1)!-1$$ Since all terms cancel out except $(n+1)!$ and $-1$

3

Here is a simple elementary proof I found, but first of all, some lemmas: This one could easily be proven by induction: $\displaystyle \prod_{k=1}^{n}\left(1+\frac{1}{k} \right) = n+1$ You can try to prove this inequality yourself since it's not difficult: $\displaystyle \left (1+\frac{1}{k} \right )^k\leq e \leq \left (1+\frac{1}{k} \right)^{k+1}\\$ This ...

3

Divide $x$ by $2$, then divide by $3$, and so on, until you cannot divide further. If you arrive at the number $1$, you started with a factorial. Otherwise you didn't.

3

Hint #1: Look at Stirling's approximation. Hint #2: $\ln n^{\ln n} = \left(e^{\ln \ln n}\right)^{\ln n} = e^{\left(\ln n\cdot \ln\ln n\right)} = ?$ (Note that I use $\ln$ rather than $\lg$, but the bases of the logs don't make any real difference here - convince yourself of that, though!)

3

Since $4!$ is congruent to $0$ (mod $12$) then any multiple of $4!$ is congruent to $0$ (mod $12$). So we need only look at the first 3 terms, and since each of the first 3 terms is congruent to themselves (mod $12$), then the addition of all the terms in (mod $12$) is: $(1! + 2! + 3! + 0 + 0 +...+ 0)$(mod 12). So the remainder should be 9.

3

$$\frac{(mk)!}{k!^m}=\frac{\color{red}{(1.2\cdots k)}\color{green}{((k+1).(k+2)\cdots(2k))}\cdots\color{blue}{(((m-1)k+1).((m-1)k+2)\cdots(mk))}}{\color{red}{k!}.\color{green}{k!}\cdots\color{blue}{k!}}$$ Does that gave you some HINT?

2

$1! + 2! + \cdots + n! \le n! + n! + \cdots + n! = n\cdot n! < (n+1)\cdot n! = (n + 1)!$

2

My other answer was answering a very different question. Looking at Wikipedia's description of fractional values in a factorial number system, it says that $e$ can be represented as $10_F1111111\ldots$, i.e. as $1\times2! +0\times 1! +1\times\frac1{2!}+1\times\frac1{3!}+1\times\frac1{4!}+1\times\frac1{5!} + \cdots$. Note that Wikipedia omits any digits ...

2


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