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25

Note that: $$e^{-1}=\sum_{k=0}^\infty \frac{(-1)^k}{k!}$$ Then: $$\frac{n!}e=n!e^{-1} = \left(\sum_{k=0}^{n} (-1)^k\frac{n!}{k!}\right) + \sum_{k=n+1}^{\infty} (-1)^{k}\frac{n!}{k!}$$ Show that if $a_n=\sum_{k=n+1}^{\infty} (-1)^{k}\frac{n!}{k!}$ then $0<|a_{n}|<1$ and $a_n>0$ if and only if $n$ is odd. So the when $n$ is odd, the value is: ...

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The number of derangements of $[n]=\{1,\ldots,n\}$ is $$d_n=n!\sum_{k=0}^n\frac{(-1)^k}{k!}\;,$$ so $$\frac{n!}e-d_n=n!\sum_{k>n}\frac{(-1)^k}{k!}\;,$$ which is less than $\frac1{n+1}$ in absolute value. Thus for $n\ge 1$, $d_n$ is the integer nearest $\frac{n!}e$, and $$d_n=\begin{cases} \left\lfloor\frac{n!}e\right\rfloor,&\text{if }n\text{ is ... 14$$ \frac{1}{\log_{2}50!} + \frac{1}{\log_{3}50!} + \frac{1}{\log_{4}50!} + \dots + \frac{1}{\log_{50}50!} =\log_{50!}2+\log_{50!}3+\log_{50!}4+...+\log_{50!}50=\log_{50!}(2\cdot3\cdot4\cdot...50)=\log_{50!}(50!)=1$$12 HINT: use that \log_2 50!=\frac{\ln(50!)}{\ln(2)} thus we get$$\frac{\ln(2)+\ln(3)+...+\ln(49)+\ln(50)}{\ln(50!)}=\frac{\ln(50!)}{\ln(50!)}=1$$10 You know n is between 23 and 28, inclusive, because there is a 23 in the prime factorization of n! but no 29. There are 4 factors of 7, so these must be from 7, 14, 21, and 28. Therefore n=28. 9$$\sum_{k=2}^{50}\frac{1}{\log_k(50!)}=\sum_{k=2}^{50}\frac{1}{\frac{\log(50!)}{\log k}}=\sum_{k=2}^{50}\frac{\log k}{\log(50!)}=\frac{1}{\log(50!)}\sum_{k=2}^{50}\log k=\frac{\log(50!)}{\log(50!)}.$$7 Following @Vladimir's comment, I can show that a=3e has this property. I don't find the proof very enlightening, though... We have$$ \frac{n!}{3e} = \sum_{k=0}^n \frac{1}{3}\frac{n!}{k!}(-1)^k + E $$where E is an error term that is small by comparison with the other terms — so it won't affect the parity of the floor except in the case where it's ... 6 The key part of this problem is to realize (1+{1\over n})^n\geq 2. We use the binomial theorem here: (1+{1\over n})^n\geq 1+n\cdot1\cdot{1\over n}=2 Now we apply our induction step: (\frac{n+1}{2})^{n+1}=(\frac{n}{2})^{n}\cdot({1+{1\over n}})^n\cdot({n+1\over2})\geq n!\cdot2\cdot ({n+1\over2})=(n+1)! 6 Hint: Bertrand's postulate(in actuality a theorem) states that for every prime p, there exists another prime number between p and 2p. This means that \forall n>1, n! will always have a single power of some prime number(s). For any positive integer n\ge 2, there exists a prime p such that \frac{n}{2}<p\le n This implies that p\mid n!. ... 5 This is a general result. We can write for any N$$\begin{align} \sum_{n=2}^N\frac{1}{\sum_{m=2}^N \log_n(m)}&=\sum_{n=2}^N\frac{1}{\sum_{m=2}^N \frac{\log_b (m)}{\log_b(n)}}\\\\ &=\frac{\sum_{n=2}^N\log_b(n)}{\sum_{m=2}^N\log_b(m)}\\\\ &=1 \end{align}$$where we used \log_n(m)=\frac{\log_b(n)}{\log_b(m)}. And we are done! 4 HINT$$ \frac{(3n)! e^n}{(2n)! n^n 8^n} = \left(\frac{e}{8n}\right)^n \prod_{k=1}^n (2n+k) = \prod_{k=1}^n \frac{2n+k}{8n/e} = \prod_{k=1}^n \left(\frac{e}{4} + \frac{ke}{8n}\right) $$4 First, that original representation you have should not have (n-2)!, but just (n-2). The "..." is just to show you are continuously multiplying by the next lowest integer until you reach 1. For someone who has never learned about the factorial that second notation could be confusing (it may seem obvious to most to stop at 1, and not 0 or ... 3 Well i can only assume what you are looking for. By the term in your question i think you might mean a superfactorial. See this Wikipedia Article on Factorials - it might help. If you mean a product of factorials it would be a notation like$$ sf(n)=\prod^n_{k=1}k!$$3 {50}\choose{4} is read "50 choose 4" and is the number of ways to choose 4 things from 50 things, where order doesn't matter. Most graphing calculators will have a button for this, if not there is the formula:$${{n}\choose{k}} =\frac{n!}{k!(n-k)!}$$where n!=n\times (n-1)\times\cdots\times 1. 3 Note that$$\frac{(2n)!}{n!n!}\leq\sum_0^{2n}\binom{2n}{k}=(1+1)^{2n}=4^n$$So$$\frac{4^n n!n!}{(2n)!}\geq 1$$2$$\frac{1}{(2m)!}\sum_{k=0}^{m}\binom{m}{k} x^k (2m-k)!=\frac{1}{(2m)!}\int_{0}^{+\infty}\sum_{k=0}^{m}\binom{m}{k} x^k z^{2m-k} e^{-z}\,dz $$and the RHS can be written as:$$ \frac{1}{(2m)!}\int_{0}^{+\infty} \left(z(x+z)\right)^m e^{-z}\,dz = \frac{e^{x/2}}{(2m)!}\int_{x/2}^{+\infty} \left(z^2-\frac{x^2}{4}\right)^m e^{-z}\,dz $$whose exact value is:$$ ...

2

Binomial Coefficient or Combination Wikipedia words it quite nicely: $\binom{n}{k}$ is read as "$n$ choose $k$", because there are $\binom{n}{k}$ ways to choose $k$ elements, disregarding their order, from a set of $n$ elements. I'd bet that most calculators use this efficient method, $$\binom{n}{k} = \prod\limits_{i=1}^k \frac{n+1-i}{i}$$ So for ...

2

You can't break apart the product like this$$\lim \sqrt[n] {n!} = (\lim \sqrt[n]{n})(\lim \sqrt[n]{n-1}) \cdots (\lim \sqrt[n]{1})$$simply because the number of terms also goes toward infinity. To illustrate this, take the related example $\lim_{n \to \infty}\sqrt[n]{n^n}$, or written out: $$1,\quad \sqrt2\cdot \sqrt2,\quad \sqrt[3]3\cdot \sqrt[3]3 \cdot ... 2 About de Polignac's formula: You found yourself how many factors 5 the number n! has: The number of factors 5 is n/5 + n/25 + n/125... If you take other prime numbers, then you get very similar results: The number of factors 2 is n/2 + n/4 + n/8 ..., then number of factors 103 is n/103 + n/103^2 + n/103^3 ... and so on. s_p(n) is the formula you found, ... 2 I think your proof is correct, but using a lemma and then doing induction is overkill for this question! In general, I would also strongly advise against trying to use both sides of an equation simultaneously in an inductive proof. It's not impossible to do, but it's one of the main sources of error when people are doing inductive proofs - you don't want to ... 2 To give an example: The beta distribution is defined via the gamma function (in the introduction of the article are some applications of this distribution mentioned). I came across this distribution, when I looked up how probabilities of binomially distributed random variables are calculated in statistical libraries. So let's have a look at scipy (a ... 2 Let N be the largest integer such that N < \xi. For n > N consider the fraction \frac{n!}{\xi^n} ans split into three parts:$$\frac{n!}{\xi^n} = \left( \frac{1}{\xi} \dotsb \frac{N}{\xi} \right) \cdot \left( \frac{N+1}{\xi} \dotsb \frac{n-1}{\xi} \right) \cdot \frac{n}{\xi}$$The first part is a fixed number S(\xi). The second part is ... 1 Let the initial limit be \ell, whose prime factorization is:$$ \ell = p_1^{e_1} \cdots p_k^{e_k} $$where p_1 < \dots < p_k. Then we are essentially taking i to be each of p_1, \ldots, p_k and iteratively dividing it out from \ell. Since i already exists in the prime factorization of \ell and there is no way that we would have divided it ... 1 If \xi is non-positive, there is nothing to prove (similar if \xi \in (0, 1]). So let's assume \xi > 1. Let n_0 = \lceil \xi \rceil. Since for each i \in \{1, \ldots, n_0\},$$\frac{n_0^2 + i}{\xi} \times \frac{i}{\xi} \geq \frac{n_0^2}{\xi^2} \geq 1,$$and \frac{m}{\xi} > 1 for all m > n_0, it follows that for all n \geq n_0^2 + ... 1 Assuming your result holds for n=k, i.e.$$k! \geqslant 2^{k-1},$$you then have$$(k+1)! = (k+1) \times k! \geqslant (k+1) \times 2^{k-1} \geqslant 2^k,$$where the first equality is the definition of the factorial, the second is the case n=k, and the third is true because k\geqslant1. 1 By the definition of double factorial, (2n-1)!! = (2n-1).(2n-3).(2n-5) \dots 1 = (2n-1).((2n-3)!!) 1 As @ADG notes in his answer,$$ \sum_{k=1}^{n}\binom{n}{k}(−1)^{n+k}k^n = \sum_{k=0}^{n}\binom{n}{k}(−1)^{n+k}k^n = \sum_{k=0}^{n}\binom{n}{k}(−1)^{k}(n-k)^n, $$since \binom{n}{n-k}=\binom{n}{k}. Now, the formular is a special case of$$ \sum_{k=0}^{n}(-1)^k\binom{n}{k}P(n-k) = n! a_n, $$where P is a polynomial of degree \leq n, and a_n is the ... 1 Let f(x) = x^3 - (x + 1)^2 f'(x) = 3x^2 - 2x + 2. f'(x) = 0 when x = \frac{2 \pm \sqrt(4 - 24}{6} so f'(x) never equals 0 for any real x. So f(x) has no extrema and thus only one real root.  f(2) = -1; f(3) = 18 so f(x) has only one real root is between 2 and 3. So f(n) > 0 iff n^3 > (n+1)^2 iff n \ge 3. ===== D'oh! ... 1 By considering the contour integral$$\oint_C d\zeta \frac{\zeta^{-z}}{(1+\zeta)^2} $$about a keyhole contour and using the residue theorem, we may derive the relation$$\left (1-e^{-i 2 \pi z} \right) \int_0^{\infty} dx \frac{x^{-z}}{(1+x)^2} = i 2 \pi \left [\frac{d}{d\zeta} e^{-z \log{\zeta}} \right ]_{\zeta=e^{i \pi}} = i 2 \pi (-z e^{-i \pi} ) e^{-i ...

1

Put "wiki factorial" into a search engine of your choice. The gamma function is the generalisation of factorial which is discontinuous at negative integers for the reason you appear to have spotted.

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