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5

$$\sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right) =\sum_{n=1}^\infty \int_0^1 \exp(u) \frac{(1-u)^{n}}{n!} du$$as everything is positive: $$\sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right)= \int_0^1 \exp(u) \sum_{n=1}^\infty \frac{(1-u)^{n}}{n!} du \\= \int_0^1 \exp(u)(\exp(1-u) - 1) du = e - \int^1_0 \exp u du = 1$$

9

The first thought that comes into my head is to write $$e - \sum_{k=0}^n \frac{1}{k!} = \sum_{k=n+1}^\infty \frac{1}{k!},$$ so that the given sum is equivalent to a double sum: \begin{align*} \sum_{n=1}^\infty \sum_{k=n+1}^\infty \frac{1}{k!} &= \sum_{k=2}^\infty \sum_{n=2}^k \frac{1}{k!} \\ &= \sum_{k=2}^\infty \frac{1}{(k-2)!k} \\ &= ... 2 As suggested in several comments, the simplest form of Stirling approximation for n! is\sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}$$(http://en.wikipedia.org/wiki/Stirling%27s_approximation) Take the logarithms and develop first$$\log\left(2n!\right) - 2\log\left(n!\right)$$which results to be (2 n+1) \log (2). The remaining is obvious. If I may, ... 5 Alternative computation:$$\begin{align*} a_n &= \frac{1}{2n} \log \binom{2n}{n} = \frac{1}{2n} \log \frac{(2n)!}{(n!)^2} \\ &= \frac{1}{2n} \sum_{k=1}^n \log \frac{n+k}{k} = \frac{1}{2n} \sum_{k=1}^n \log \Bigl( 1 + \frac{1}{k/n} \Bigr). \end{align*}$$Therefore, as n \to \infty, we get a Riemann sum:$$\begin{align*} \lim_{n \to \infty} a_n ...

6

One easy way is to observe in the binomial expansion of $(1+1)^{2n} = \sum\limits_{k=0}^{2n} \binom{2n}{k}$, the term $\binom{2n}{n}$ is the largest one among all the $2n+1$ terms of the form $\binom{2n}{k}$. This gives us a bound $$\frac{2^{2n}}{2n+1} \le \binom{2n}{n} \le 2^{2n} \quad\implies\quad \log 2 - \frac{\log(2n+1)}{2n} \le ... 0 If b|n then the quotient n/b is odd only if b contains the largest factors of two in each of 2,4,...,|n/2_| , where |n/2_| is the largest even number smaller than n. Otherwise, b|n is even. For example, for n=10, then, n/b is odd (assuming b divides 10!) , if b contains each of the largest factors of 2 in each of 2,4,6,8,10, which ... 9 There's a mistake in your first attempt:$$ x = \frac{10!}{8!} + \frac{10!}{9!} = 10 \times 9 + 10 = 100 0 Note that N!=2^ak where k is some odd integer and a is the highest power of 2 that divides N! which is computed by the number of times 2 appears + the number of times 4 appears + ... until zero times for that power of 2. For example, 4! = 4*3*2*1 would have a couple of 2 factors and one 4 factor thus, 8 divides 4! as the greatest power of 2 here. ... 0 Since M can be much larger than N, you need Stirling's approximation in many cases but with more correction terms than are usual. The exact number of factors of 2 in N! are not useful in most cases. 0 The numbers you can construct are of the form \sum_{j=0}^nc_jj! with c_j\in\{-1,\,0,\,+1\}. Imagine you want to construct a number smaller than 4165. Since 7!-\sum_{j=0}^6j!=4166, you can only use the numbers k! with 0\leq k\leq 6. That's 7 numbers. Using your formula you can construct 3^7=2187 numbers this way. Sadly, 2187<4165 so the ... 1 We show that  \lim_{x \to \infty} \frac{\sum{1 \leq i \leq n}log(i)}{n log(a)} = \infty. Indeed, \sum_{1 \leq i \leq n}log(i) > \sum_{n/2 \leq i \leq n}log(i). Note that for all i \geq n/2, we have log(i) \geq log(n/2) = log(n)-1. Hence, we have \sum_{n/2 \leq i \leq n}log(i) \geq \frac{n}{2}log(n) - \frac{n}{2}. Therefore, \sum_{1 \leq i \leq ... 10 Note that \begin{align} \log a_n&=\log \sqrt{1!\sqrt{2!\cdots\sqrt{n!}}}=\frac{1}{2}\log 1! +\frac{1}{4}\log 2!+\cdots+\frac{1}{2^n}\log n! \\ &=\sum_{k=1}^n \frac{\log (k!)}{2^k}=\sum_{k=1}^n\frac{1}{2^k}\sum_{j=1}^k\log j= \sum_{k=1}^n \log k \Big(\sum_{j=k}^n \frac{1}{2^j}\Big). \end{align} Hence, then sequence \log a_n, which is increasing ... 0 I come from a background in computers, so here's my two cents. Taking the logarithm to the base 10 of n!. If the log comes out to be x, it is not hard to see that the number of digits must be the lowest integer greater than or equal to x, i.e, floor(x)+1. Now the question comes down to approximating the log(n!) It is possible to prove by induction that ... 7 Hint: If n\geq 5, what is the last digit of n! ? 0 Hint: for the first one, you are looking for two factorials that have almost the same terms, but the second has (k+1)(k+2) in addition to what the first has. Dividing them will give the desired result because the matching terms will cancel. The second doesn't have as clean a representation, but you can modify the first to get something. 0 \frac{1}{(k+1)(k+2)}=\frac{1}{k+1}-\frac{1}{k+2} \sum_{k=1}^n\frac{1}{(k+1)(k+2)}=\sum_{k=1}^n(\frac{1}{k+1}-\frac{1}{k+2}) (\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\ldots +(\frac{1}{n+1}-\frac{1}{n+2}) =\frac{1}{2}-\frac{1}{n+2}=\frac{n}{2(n+2)} 2 There is an important self-similar sequence that goes: 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, where I've grouped terms into bigger blocks. The "pattern" that makes this self similar is "=, =, =, =, +1" where "=" means repeat the working block unit and "+1" means to increment the block unit by adding one to the last ... 3 For k \geqslant 2, we havef(5^k) = \frac{5^k-1}{5-1} \equiv 1 \pmod{5},$$so the multiple$$\frac{5^{k-1}-1}{4}\cdot 5$$of 5 is skipped. That produces the sequence 5,5+5^2 = 30, 5+5^2+5^3 = 155, 155+5^4 = 780,\dotsc of skipped multiples of 5. There are more, for example f(350) = 70+14+2 = 86, so 85 is skipped. Generally, if we call ... 1 Using the limit formula: http://www.sms.edu.pk/journals/jprm/jprmvol8/01.pdf. Using the integral formula: http://www.math.unl.edu/~sdunbar1/ProbabilityTheory/Lessons/StirlingsFormula/GammaFunction/gammafunction.pdf. 1 The matrix of Eulerian-numbers might help here. There are two (only slightly) different definitions; let's use that of the matrix which begins with$$ E=\small \begin{bmatrix} 1 & . & . & . & . & . \\ 1 & 0 & . & . & . & . \\ 1 & 1 & 0 & . & . & . \\ 1 & 4 & 1 & 0 & . & . ...

4

Long time ago, needing an explicit formula for statistical thermodynamics, I was facing the same problem of the calculation of $n!$ for $0<n<1$. Starting from Ramanujan's approximation, what I did was to write $$n!\approx\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+x(n)}$$ What can be easily established is that $$x(0)=\frac{1}{\pi ... 5 Interestingly, if we do a series expansion in Mathematica Series[(Gamma[n + 1]/Sqrt[Pi] (n/Exp[1])^-n)^6, {n, Infinity, 6}] We get$$8 n^3+4 n^2+n+\frac{1}{30}-\frac{11}{240 n}+\frac{79}{3360n^2}+\frac{3539}{201600 n^3}-\frac{9511}{403200 n^4}-\frac{10051}{716800n^5}+\frac{233934691}{6386688000n^6}+O\left(\frac{1}{n}\right)^{13/2}.$$This suggests that ... 4 The fraction \tfrac{1}{30} may not be, practically speaking, the best upper bound for the final term of the radicand. Indeed, according to Ramanujan's own notes, the number ranges from \tfrac{1}{30} down to \tfrac{1}{100}. Observe that when evaluated at 0, ... 2 Setting$$ a_n=\frac{n^n}{n!}, $$we have$$ \frac{a_{n+1}}{a_n}=\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}=\frac{(n+1)^n\cdot n!}{n!\cdot n^n}=\frac{(n+1)^n}{n^n}=\left(1+\frac{1}{n}\right)^n \quad \forall n. $$Since$$ \lim_n\frac{a_{n+1}}{a_n}=\lim_n\left(1+\frac{1}{n}\right)^n=e>2, $$there is an N \in \mathbb{N} such that$$ ...

0

check this out: http://en.wikipedia.org/wiki/Stirling%27s_approximation This is a famous estimation of n!.

0

The result is clear if $|a|<1$. If $|a|>1$, I will now show the calculation with $a>1$ (it's no different with $a<-1$. The $n$-th value of the sequence equals $$\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdot\frac{a}{4}\cdots\frac{a}{n}.$$ Now, let $N$ be the integer which is larger than $a^2$. Let $C=\frac{a^N}{N!}$. Now take any integer $k$ ...

1

Compare it with the following geometric sequence: $b_n=(\frac{a^m}{m!})(\frac{a}{m+1})^n,$ where $m$ is the smallest positive integer such that $m+1\gt a.$ Notice that $a_{n+m}\le b_n$ so that $\lim a_n=\lim b_n=0.$ Hope this helps.

2

There can't be any multiplicative formula of the sort you describe for $4n\choose 2n$ in terms of $2n\choose n$, because there are prime factors of the former that aren't factors of the latter. By Bertrand's Postulate there's a prime number between $k=2n$ and $2k$($=4n$); that prime will be a factor of $4n\choose 2n$, but can't be a factor of $2n\choose n$ ...

4

Here is an estimate that gives a good approximation of $\binom{4n}{2n}$ in terms of $\binom{2n}{n}$. Using the identity $$(2n-1)!!=\frac{(2n)!}{2^nn!}\tag{1}$$ it is straightforward to show that $$\frac{\binom{4n}{2n}}{\binom{2n}{n}}=\frac{(4n-1)!!}{(2n-1)!!^2}\tag{2}$$ Notice that \begin{align} \frac{(2n-1)!!}{2^nn!} ... 2 I think this is what you are looking for (but this is just back substitution):\binom{4n}{2n} = \frac{(4n)!}{((2n)!)^2} = \frac{(4n)!}{\binom{2n}{n}^2 (n!)^4}.$$Please update me on whether this is what you're looking for... 5 "Higher order of growth" does not mean that n!\lt n^n but the stronger property that$$n!/n^n\to0.$$To prove that this property holds, note that n\geqslant2k for every 1\leqslant k\leqslant n/2 and n\geqslant k for every n/2\lt k\leqslant n, hence$$ n^n=\prod_{k=1}^{n}n\geqslant\prod_{1\leqslant k\leqslant n/2}(2k)\cdot\prod_{n/2\lt k\leqslant ...

2

If $n\in \mathbb{N}$, Then $n! = 1 \times 2 \times 3\times 4...............\times (n-2)\times (n-1)\times n$ Now $n\geq n$ and $n>(n-1)$ and $n>(n-2)$ and $n>(n-3)$......... $n>4\;\;,n>3\;\;,n>2\;\;,n>1$ So $n\cdot n\cdot n\cdot ............\cdot n(n-\bf{times})\geq n\cdot (n-1)\cdot (n-2)....3\cdot 2 \cdot 1$ So $n^n\geq n!$

1

Your professor is wrong, and the explanation can be found here on page 5.

3

You are right that $n^n$ grows faster than $n!$. You can see this question with answer (thanks to @JohnHabert for his comment above) for a proof which says exactly what you are saying. This is illustrated by just considering the first couple of numbers \begin{align} 1^1 = 1 \quad &\quad\quad 1! = 1\\ 2^2 = 4 \quad &\quad\quad 2! = 2\\ 3^3 = 27 ... 1 Hint: Consider the binomial expansion of (1+1)^{n-2}. 0 this is not a question of a recurrence relation, instead its a question of divide and conquer technique to find the factorial, which wont be useful btw, complexity will remain same, as the numbers are eventually have to be multiplied. (This is not the same as merge sort, in merge sort one can gain advantage because of it is easier to sort already sorted ... 1 If you wish to make an inverse of the factorial for a fractional value, try starting with one of the aproximations here and solving for x. The two formulas arex!\approx\sqrt{2\pi}x^xe^{-x}\sqrt{x+\frac{1}{6}+\frac{1}{72x}-\frac{31}{6480x^2}-\frac{139}{155520x^3}+\frac{9871}{6531840x^4}}\\ ...

0

Your 3-horse Exacta would be ${3 \choose 2}\frac{2!4!}{6!}$. You choose two of the three horses to be the two that win, then the same calculation you did for the 2-horse Exacta. You are correct that this is $\frac 15$ If your 4-horse Trifecta is choose four and three of them have to be the top three, that is ${4 \choose 3}\frac{3!3!}{6!}=\frac 15$ I ...

11

It should be pretty obvious why $10!$ and all higher factorial must all have at least one zero at the end: they're all divisible by $10$. $$10! = \mathbf{10} \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$$ If you think about it a bit more, it's also pretty obvious that $20!$ and any factorials above it must haveat least ...

0

Immediately we see that any $n$ that's a power of 2 should work. e.g. $4! = 3!2!2!$, $8! = 7!2!2!2!$, $16! = 15!2!2!2!2!$ Similarly, by combining whatever small factorials we want, we can take $n!$ and $(n-1)!$ to have that ratio. For instance, suppose we wanted to use $3!5!7! = 3628800$ somewhere; we can then make $3628800! = 3628799!7!5!3!$. You can ...

0

No. Consider $4^4 = 2^{8}$, which has no 3 in its prime factorization, while $4! = 1 \cdot 2 \cdot 3 \cdot 4 = 2^3 \cdot 3$.

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