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4

One approach for calculating the limit as $n \to \infty$ is to use the Euler-Maclaurin sum formula twice. We'll begin by rewriting $$ \begin{align} \frac{1}{n^2} \sum_{i=0}^{n} \log \binom{n}{i} &= \frac{1}{n^2}\left[(n+1)\log n!-2\sum_{i=0}^{n} \log i!\right] \\ &= \frac{(n+1)\log n!}{n^2} - \frac{2}{n^2}\sum_{i=0}^{n} \log i!. \tag{1} \end{align} ...


4

Matrix multiplication is not commutative, so (maybe) that symbol means that you are considering the product adding the next matrix to the right: $$ X_1 X_2 \cdots X_k. $$ Anyway, I am just guessing, I've never seen it before!


4

Below I've left my previous flawed approach. It is interesting that my upvoters and I hadn't noticed the highlighted mistake. And also that there exists a very straightforward proof: $$\prod_{m=1}^n\frac{p_m}{p_m-1}-\ln n>\sum_{m=1}^n\frac{1}{m}-\ln n>\gamma>0. $$ If you wish to avoid Mertens but not Rosser, note that your inequality is weaker ...


4

We have $$\prod_{i\leq n}\frac{p_{i}-1}{p_{i}}=\prod_{i\leq n}\left(1-\frac{1}{p_{i}}\right)=\frac{1}{\log\left(p_{n}\right)e^{\gamma}}+O\left(\frac{1}{\log^{2}\left(p_{n}\right)}\right) $$ by the Mertens theorem. Now note that, by Rosser's theorem ...


3

Let $S(n)$ be the sum defined by $$S(n)\equiv\frac{1}{n^2}\sum_{k=0}^{n}\log \binom{n}{k} \tag1$$ Expanding terms in $(1)$ yields $$\begin{align} S(n)&=\frac{1}{n^2}\left((n+1)\log n!-2\sum_{k=0}^{n}\log k!\right)\\\\ &=\frac{1}{n^2}\left((n-1)\log n!-2\sum_{k=2}^{n-1}\log k!\right)\tag2 \end{align}$$ Substituting $\log k!=\sum_{\ell=2}^{k}\log ...


3

You have made the problem much hard than it is suppose to be. For the left hand side, simply observe that $\frac{1}{2}>\frac{1}{3}$ and $\frac{3}{4}>\frac{3}{5}$ and so on... to $\frac{1997}{1998}>\frac{1997}{1999}$. Now, take the product of these inequalities to get the desired result. For the other inequality, it is a similar trick except you ...


3

The (Gauss) $q$-analogue of the binomial theorem expresses the product of terms $(1 - q^i)$ as a sum whose terms are (up to some normalization factors) $q$-binomial coefficients. Formulas are listed at Gaussian binomial coefficients in Wikipedia, for example.


3

$$\begin{align} \prod_{n=1}^N\frac{m+1}{m}\frac{m+x}{m+x+1}&=\left(\prod_{n=1}^N\frac{m+1}{m}\right)\left(\prod_{n=1}^N\frac{m+x}{m+x+1}\right)\\\\ &=\left(\frac{2}{1}\frac{3}{2}\frac{4}{3}\cdots \frac{N}{N-1}\frac{N+1}{N}\right)\left(\frac{1+x}{2+x}\frac{2+x}{3+x}\frac{3+x}{4+x}\cdots \frac{N-1+X}{N+x}\frac{N+x}{N+1+X}\right)\\\\ ...


2

In terms of known notations: Let $u_{n+1} = 2 \, u_{n} + 1$ where $u_{0}=1$, for which $u_{n} = 2^{n}-1$. Using $$(x;q)_{n} = \prod_{k=0}^{n-1} (1 - x \, q^{r})$$ leads to \begin{align} P_{n} &= \prod_{k=0}^{n-1} \{ u_{k} \} = \prod_{k=1}^{n} \{ 2^{k+1} - 1 \} \\ &= 2^{\binom{n+1}{2}} \, \prod_{k=1}^{n} \left(1 - \frac{1}{2^{k}}\right) \\ &= ...


2

Suppose that $\neq:X\times X\to S$ is continuous. Let $E\subseteq X\times X$ be defined as $\neq^{-1}(\{\top\})$, the inverse image of $\{\top\}$ under the function $\neq$. Since $\neq$ is continuous and $\{\top\}$ is open in $S$, if follows that $E$ is open in the product topology. In fact, $E=\{(x,y)\in X\times X\,|\,x\neq y\}$ by the definition of $\neq$. ...


1

$${ P }_{ n }^{ 2 }\left( 999 \right) =\frac { { 1 }^{ 2 }\cdot { 3 }^{ 2 }...\cdot { \left( 2\cdot 999-1 \right) }^{ 2 } }{ { 2 }^{ 2 }\cdot { 4 }^{ 2 }...\cdot { \left( 2\cdot 999 \right) }^{ 2 } } =\frac { 1\cdot 3 }{ { 2 }^{ 2 } } \cdot \frac { 3\cdot 5 }{ { 4 }^{ 2 } } \cdot ...\cdot \frac { \left( 2\cdot 999-1 \right) \left( 2\cdot 999+1 \right) }{ ...


1

Yes, although it is a capital letter $C.$ Back on page 65, formula (2.9) is followed by where $C$ is Euler's constant. The same is used in the integral table book of Gradshteyn and Ryzhik, especially in section 8.36, pages 952-956 in the fifth edition


1

Regarding the empty product you're right. In case the lower index of a product is greater than the upper index, the product is equal to one (analogously to the empty sum which is zero in such cases). We can read e.g. in H. Bateman, Higher Transcendental Function Volume 1, section 5.3, p.207 Definition of $G$-Function: ... where an empty product ...


1

Your "ternary dot product" is basically equivalent to the Hadamard product (https://en.wikipedia.org/wiki/Hadamard_product_%28matrices%29): given two vectors $\langle a_i\rangle, \langle b_i\rangle$, their Hadamard product is just componentwise multiplication: $$\langle a_i\rangle*\langle b_i\rangle=\langle a_ib_i\rangle.$$ To see why this is equivalent to ...


1

You could call it, for example, ${\bf x}^T Y {\bf z}$ where $Y$ is the diagonal matrix with the entries of ${\bf y}$ on the diagonal. Of course you could apply any permutation to ${\bf x}$, ${\bf y}$, ${\bf z}$.


1

If you write out the first few terms of the product, you'll see that most of the factors cancel out: $$ \require{cancel} \begin{align} \prod_{m=1}^{\infty} \frac{m+1}{m}\cdot\frac{m+x}{m+x+1} &= ...



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