# Tag Info

24

Let $x$ be a real number such that $x>0$. One may observe that, using successive integrations by parts, we have $$\int_0^n t^{x-1} \left( 1-\frac{t}{n}\right)^n{\rm{d}} t= \frac{n! \:n^x}{x(x+1)(x+2)\cdots(x+n)},\quad n=1,2,\ldots,$$ leading to $$\Gamma(x)=\lim_{n\to+\infty}\left(\frac{n! \:n^x}{x(x+1)(x+2)\cdots(x+n)}\right).$$ Then, as $n$ is ...

19

No, for odd $n,$ there is a prime between $(n-1)/2$ and $n.$ The exponent of this prime in factoring $n!!$ is one, that is, odd. Edit. looking at even numbers and the definition, it appears $$(2n)!! = 2^n n!$$ in which case we ignore the exponent of $2$ and concentrate on $n!,$ which cannot be a square either for this $n \geq 3,$ also because of an odd ...

11

Hint. If $x$ was $1$, this would be asking for an equivalent of $n!$, so you'd need Stirling's formula. In general, the function you've written is equal to $\frac{1}{\Gamma(x)} \Gamma(x + n + 1)$, so you can still obtain an equivalent directly from Stirling's formula, which is also applicable to the Gamma function.

8

One way to do this is to notice that: $$(n+t)!\approx n!n^t$$ when $n$ is large. For example, we have $(n+3)!=n!(n+1)(n+2)(n+3)\approx n!n^3$. Also notice that: $$t!=\frac{(t+n)!}{(t+1)(t+2)\dotsb(t+n)}$$ Putting these together, we have: $$t!=\lim_{n\to\infty}\frac{n!n^t}{(t+1)(t+2)\dotsb(t+n)}$$ which is valid even when $t$ is not an integer! In case you ...

7

Let $f(n)=[\prod_{i=0}^n (1+r/n)]^{1/n}$ Then, $\ln f(n)=1/n\sum_{r=0}^n\ln (1+r/n)\Rightarrow \lim_{n\to \infty}\ln f(n)=\int_{0}^1 \ln (x+1) dx=\ln 2-\int_{0}^1 \dfrac{x}{x+1}dx=2\ln 2-1\Rightarrow \lim_{n\to \infty} f(n) =4/e$

7

Here is a combinatorial proof: Arrange $n^2$ person in $n^2$ chairs in a row, we have $n^2!$ ways. Alternatively, we can pick $n$ persons for the first $n$ chairs, then $n$ persons for chairs $n+1$ to $2n$, then $n$ persons for $2n + 1$ to $3n$, etc, then arrange each group of $n$ persons, we have $\left(\prod_{k=0}^{n-1}{n^2-kn \choose n}\right)(n!)^n$ ...

6

Hint: $$(n+1)!+(n+1)(n+1)!=(n+1+1)[(n+1)!]=(n+2)(n+1)!=(n+2)!$$

4

By the ratio test $$\frac{(2n+1)!}{(2n-1)!}\frac{n^n}{(n+1)^{n+1}}\sim4n\left(1+\frac1n\right)^{-n}\sim_\infty\frac4en\to\infty$$ so the given limit is $+\infty$.

3

Yes you're right and we have using the same idea $$P(A\cap B)=\frac{C_8^8\times C_{44}^5}{C_{52}^{13}}$$

3

Let $n=2m$, and rewrite the identity as $$2^mm!=\frac{(2m)!}{(2m-1)!!}\;.\tag{1}$$ Now notice both sides of $(1)$ are simply $$\prod_{k=1}^m2k\;,$$ since cancellation removes the odd factors in the numerator on the righthand side.

3

We have: k! ~ ((k/e)^k)*square(2/pik) (stirling) equivalence is compatible with ln, that is: an ~bn, an,bn>0 => ln(an) ~ln(bn) we get: ln(k!) ~ k*ln(k) (I skipped the calculus) ln(ln([k!]^k!)) = ln(k!) + ln(ln(k!)) ~ ln(k!) (ln(x)/x -->0, when x goes to infinity) Let Vk be the general term of your series: Vk ~ 1/ln(k!) ~ 1/(k*ln(k)) = Uk The ...

3

Here's a quick solution to this in R $($although it depends on floating-point arithmetic$)$: n = log(1:100000, 10) logf = cumsum(n) first.four = floor(10^(logf-floor(logf))*1000) which(first.four == 1999) The output is 15998 19796 20030 20678 24284 25809 28019 28956 30752 31432 33289 38840 51822 52487 53962 56006 56660 59986 69481 ...

3

We shall use that: Fact. $\displaystyle\lim_{n\to\infty}\left(\frac{n^n}{n!}\right)^{\!1/n} \!\!\!\!\!\longrightarrow\mathrm{e}.$ Then $$\left(\frac{(2n)!}{n^n n!}\right)^{1/n}=\left(\frac{2^{2n}\frac{n^n}{n!}}{\frac{(2n)^{2n}}{(2n)!}}\right)^{\!1/n}=4\frac{\left(\frac{n^n}{n!}\right)^{1/n}}{\left(\frac{(2n)^{2n}}{(2n)!}\right)^{1/n}} ... 3 You can use gamma function with the property that \Gamma(n+1)=n! in particular look here http://en.wikipedia.org/wiki/Gamma_function. \Gamma(4.6)=3.6!=13.383.... 3 The factorial function n! = \prod\limits_{i=1}^n i is defined only for positive integers n. But a standard move in function theory is to extend such functions to the reals, and then the complex plane. To do this, you have to decide what the logical "generalization" of the factorial function would be. An obvious place to start is the observation that ... 3$$\frac{n!}{k!(n-k)!} = \frac{n(n-1)(n-2) \cdots (n-k+1) (n-k)!}{k!(n-k)!} = \frac{n(n-1)(n-2) \cdots (n-k+1)}{k!}$$For instance, if n=7 and k=4, then$$\displaystyle{7 \choose 4} = \frac{7!}{4!\cdot(7-4)!} = \frac{7!}{4!\cdot3!} = \frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{4!\cdot3\cdot2\cdot1} = \frac{7\cdot6\cdot5\cdot4}{4!}.$$3 The sum of the binomial coefficients is in fact 2^n. What you quoted was the general formula for the binomial coefficient. You need to multiply top and bottom by (n-k)!, from which the result comes immediately. 3 Try the numbers:$$ k!+2,\,k!+3,\ldots,k!+k $$and you have k-1 consecutive composite integers. 2 Considering$$S=\sum_{k=1}^{\infty}\frac{[\lambda(k-1)+\theta][\lambda(k-2)+\theta][\lambda(k-3)+\theta]...[\lambda+\theta]\theta}{(k!)^2\mu^k}$$and using a CAS, the numerator in the summation is$$\theta\prod_{i=1}^{k-1} (\lambda i +\theta)=\theta \lambda ^{k-1} \left(\frac{\theta +\lambda }{\lambda }\right)_{k-1}$$where appears the Pochhammer symbol. ... 2 Using a computer^* we can find that$$15998! = 1999638128545238518 \dots$$(^*) Haskell one-liner: find (\(a,xs) -> take 4 (show xs) == "1999")  zip [0..] (scanl (*) 1 [1..]) 2 Suppose you have a phrase composed of n letters, with n_i letters of each kind, so that \sum_{i=1}^s n_i=n. We can arrange this in M=\binom{n}{n_1,\ldots,n_s} ways. Collapsing the block of n_i letters into 1 single block of adjacent letters, gives n-n_i+1 letters, and this can be arranged in N=\binom{n-n_i+1}{n_1,\ldots,1,\ldots,n_s} ways. ... 2 It holds for n=1. Assume that it holds for n=k, i.e.$$a(k)=k!\left(\frac{1}{0!}+\frac{1}{1!}+\cdots+\frac{1}{(k-1)!}\right).$$Then, we have ... 2 Let n = 2k, then: LHS = \dfrac{k!}{(2k)!}, and RHS = \dfrac{1}{2^k\cdot (2k-1)!!} = \dfrac{k!}{(2k)!!(2k-1)!!} = \dfrac{k!}{(2k)!} = LHS. So it is a true statement. 2 Let's try induction. For 0\le n< p we have k(n)=0 and also \frac1{p-1}(n-s(n)_p)=0. For n\ge p we can write n=mp+r with m\in\mathbb N and 0\le r<p and may assume that k(m)=\frac1{p-1}(m-s(m)_p). We note that s(n)_p=s(m)_p+r. The multiples of p that are \le n are of the form pt with 1\le t\le m and the contribution of pt to ... 2 Here's how to attempt this problem. Assume that the S and the Y are always together, meaning they form "one" letter. Now we have that there are 11 letters. Also, note that there is only technically 1 Y, since the "letter" SY is included now. Now, we have that:$$\frac{11!}{3!1!} $$2 You're right about subtracting a term; in fact, there's a (clever) strategy called "telescoping sums" and it's particularly useful here, and you won't need induction to show it. You want terms to cancel out so that you're left with the first and last terms only. If you want to do it yourself, then stop reading here and meditate on this idea: how can you ... 2 L = \lim_{n \to +\infty} \frac{a_{n+1}}{a_n}. L is infinite therefore the limit of a_nis infinite. 1 We can write the above relation as below: \sum_{k=1}^{n}k.k!=\sum_{k=1}^{n}(k+1-1)k!=\sum_{k=1}^{n}(k+1)!-\sum_{k=1}^{n}k!=\sum_{k=2}^{n+1}k!-\sum_{k=1}^{n}k!=\sum_{k=1}^{n+1}k!-\sum_{k=1}^{n}k!-1=(n+1)!-1 1 I overlooked the condition that we're supposed to read a page per day. In that case, we only need 19 squares:$$\boxed{\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square}$$Now we write down that we're going to read 4,6,10 as follows: ... 1 By the Bernoulli inequality, the sequence given by a_n=(1+1/n)^n is increasing, hence the first inequality is trivial. The second one follows from the binomial theorem:$$\left(1+\frac{1}{n}\right)^n = \sum_{j=0}^{n}\binom{n}{j}\frac{1}{n^j}=1+\sum_{k=1}^{n}\frac{\frac{n}{n}\cdot\frac{n-1}{n}\cdot\ldots\cdot\frac{n-k+1}{n}}{k!}\leq ...

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