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6

$$\begin{align} P &=\large\prod_{n=1}^\infty e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}\\ &=\large\lim_{m\to\infty}\prod_{n=1}^m e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}\\ &=\large\lim_{m\to\infty}e^m\left[\left(\frac 12\right)^1\left(\frac 23\right)^2\left(\frac 34\right)^3\cdots \left(\frac m{m+1}\right)^m\right] ...


6

Note that if $\displaystyle\prod_{i=1}^n a_i=a$ then $\displaystyle\prod_{i=1}^{n-1}a_i = \frac a{a_n}$. Now note that $\displaystyle\prod_{i=1}^1 a=a$, so $\displaystyle\prod_{i=1}^0a=1$.


4

We have: $$\log P = \sum_{n=1}^{+\infty}\left(1+\left(n+\frac{1}{2}\right)\log\left(1-\frac{1}{n+1}\right)\right)$$ but: $$\sum_{n=1}^{N}\left(1+n\log n-(n+1)\log(n+1)\right) = N-(N+1)\log(N+1)$$ since we have a telescopic sum, while: $$\sum_{n=1}^{N}\left(\frac{1}{2}\log n+\frac{1}{2}\log(n+1)\right)=\log(N!)+\frac{1}{2}\log(N+1)$$ so: ...


4

Since all the $a_k \lt 1$ (except at one point), I would try expanding $$\frac{1}{1-\prod_k^n a_k}=1+\left(\prod_k^n a_k \right)+\left(\prod_k^n a_k \right)^2+\cdots$$ (Geometric series) This will allow you to turn your problem into a sum of integrals in which, in every term of the sum, the $a_k$'s appear multiplicatively and thus can be integrated ...


3

The empty sum is zero. The empty product is one. Note that we have $0+x = x$ for all $x$ and $1\cdot x=x$ for all $x$, so that these are the corresponding values that doesn't change the final sum/product.


3

Alex R. and Asaf Karagila pointed out that it fits well with the usual identities. This might be a more structural refinement of the argument: For any set of factors $A$ you have the identity $A ∪ ∅ = A$. You would like to have $\prod_{a ∈ A} f_a · \prod_{b ∈ B} f_b = \prod_{x ∈ A ∪ B} f_x$. You need this all the time for usual reasoning with splitting up ...


3

We can take logarithms, then the inequality becomes $$\sum_{i=1}^n \log (1-a_i) \leqslant \log c.\tag{1}$$ Expanding the logarithms in Taylor series, $$\log (1-x) = -\sum_{k=1}^\infty \frac{x^k}{k}$$ and multiplying with $-1$, it becomes $$\sum_{i=1}^n \sum_{k=1}^\infty \frac{a_i^k}{k} \geqslant -\log c.\tag{2}$$ Since all terms on the left are ...


3

Using the suggested factorizations, and using $\begin{array}\\ k^2-k+1 &=k(k-1)+1\\ &=(k-1+1)(k-1)+1\\ &=(k-1)^2+(k-1)+1\\ \end{array} $ (this is really the key), $\begin{array}\\ \prod_{k=2}^n \dfrac{k^3-1}{k^3+1} &=\prod_{k=2}^n \dfrac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)}\\ &=\dfrac{\prod_{k=2}^n (k-1)}{\prod_{k=2}^n ...


2

It is a special case of the Brunn-Minkowski Theorem. It isn't strange at all; it is a rather useful inequality in measure theory.


2

About the proof: is it clear why $M\otimes_R R/J\simeq M/JM$? If yes, then consider $M\otimes_R N=0$ and tensor it (over $R$) with $R/J$ left and right. Then you get $M/JM\otimes_{R/J} N/JN=0$. Thus you have two vector spaces, namely $M/JM$ and $N/JN$, whose tensor product is $0$. But non-zero vector spaces have bases, so their tensor product, and then we ...


2

Every product should start with $1$. For instance $$\prod_{i=1}^2a_i:=1\cdot a_1\cdot a_2$$ This mode of thinking at least makes the empty product naturally work out to $1$. But it's more than that. If you think of the above in the usual way as $a_1\cdot a_2$,then it's not really symmetric. It reads that you started with the object $a_1$, and then brought in ...


2

I see your vague question and raise you a vague answer. The punchline is that you should think of $X^{[0, 1]}$ as being analogous, not to $X \times X$, but to $X$. If $f : Y \to X$ and $g : Y \to X$ are two maps, consider the following two diagrams. First, consider the diagram $$Y \xrightarrow{(f, g)} X \times X \xleftarrow{\Delta} X$$ where $\Delta$ is ...


2

Hint: $$\langle x, y\rangle \langle w, z\rangle = \langle \langle x, y\rangle w, z\rangle $$


1

Of course the product, which is after all a scalar (call it $S$) , can be written as an inner product of two vectors, e.g., $\vec{n} = (S, 0, 0, \ldots), \vec{m} = (1, 0, 0, \ldots)$. But the decomposition of that scalar is not unique, and in fact, you can find an infinite number of correct $\vec{m} $ cevtors for any specified non-zero $\vec{n}$.


1

The proof is right, so long as neither $A$ nor $B$ are empty (e.g. if $B$ is empty, we can choose $A$ freely, so the theorem fails; it could be that $A\not\subseteq C$) - you implicitly assume this when you take an element from each.


1

Assuming $X$ is an integer $\geq3$ your test is fine from a logical point of view. But it is completely useless if you want to find out whether $X:=63\,176\,591$ is prime. You would need millions of terms of the sine series in order to compute a single factor of your product, say $\sin\bigl(\pi{X\over 31}\bigr)$. But even if you computed all these factors ...


1

Yes, it's possible to find such vector, in your example the answer could by (for example): $$m=<u,v>x, n=y$$ or $$m=<x,y>u, n=v$$ So it isn't unique.


1

The only nontrivial part is to show that $S$ is closed under inversion. But since $S$ is finite, every element of $S$ has finite order. For a counterexample in the infinite case, take $G = \mathbb{Z}$ and $S = \mathbb{Z}^{\geq 0}$.


1

The crucial step is that $$\begin{align} \prod_{i=1}^{n+1}(1+a_i)&=(1+a_{n+1})\prod_{i=1}^{n}(1+a_i)\\&=\prod_{i=1}^{n}(1+a_i)+a_{n+1}\prod_{i=1}^{n}(1+a_i)\\&\ge 1+\sum_{i=1}^na_i + a_{n+1}\cdot 1\\&= 1+\sum_{i=1}^{n+1}a_i\end{align}$$


1

Hint: $a_{n,i}$ are the Stirling numbers of the first kind


1

An empty sum is defined to be $0$; an empty product is defined to be $1$. It's the same thing. For more concrete motivation, suppose $S$ is a finite set of integers, and $n$ is an integer not in $S$. Then we want the following to hold, in general: $$\prod_{i\in S \cup \{n\}} i = n \cdot \prod_{i\in S}i$$ But if $S$ is the empty set, then this equation ...


1

It's a definition that respects breaking down the product into pieces, just like saying $1=1+0$. For example, you have that $\prod_{j=1}^n a_j = (\prod_{j=1}^n a_j)(\prod_{j\in\{\phi\}}a_j)=\prod_{j=1}^n a_j$. Similarly, the empty sum $\sum_{2<j<1}a_j=0$ so that linaerity is preserved: $\sum_{j=1}^n a_j=\sum_{j=1}^na_j+\sum_{j\in\{\phi\}}a_j$. This is ...


1

This is Mahler's inequality. We had it here in this question (see also the linked questions), and it was question A2 on the 2003 Putnam, so you can find some proofs in Kedlaya's archive. As for its usage: It is sometimes used to prove the Brunn-Minkowski inequality (of which it is a special case, as Umberto P. noted). In that proof, one first proves this ...



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