proof of convergence for the polygon circumscribing constant $\sum _{3} ^{\infty} \ln(\sec(\pi/ n))$

The polygon circumscribing constant is found by: $ $ \prod _3 ^\infty \sec \left( \frac\pi n \right)$ $

I am trying to find a proof that this product converges. I know it is equal to:

$ $ \exp \left( \sum _{3} ^{\infty} \ln\left(\sec\left(\frac\pi n\right)\right) \right)$ $

So I just need to show that sum converges. I do not see an easy way to use any of the convergence tests. In particular, I spent way to long trying to integrate this function to no avail.

So what convergence test is useable in this case?

Limits: factoring out x from $\lim _{x\to +\infty }\left(\frac{5-x^3}{8x+2}\right)$

So my teacher said that I cannot use arithmetic operation to factor out x from this type of equation, saying that it’s because it’s composed only by addition and subtraction. But I don’t understand clearly, because I get the right answer (according to the book):

$ \lim _{x\to +\infty }\left(\frac{5-x^3}{8x+2}\right) =\lim _{x\to \infty \:}\frac{x×\left(\frac{5}{x}-x^2\right)}{x×\left(8+\frac{2}{x}\right)}=\lim _{x\to \infty \:}\frac{\frac{5}{x}-x^2}{8+\frac{2}{x}}=\frac{\lim _{x\to \infty \:}\left(\frac{5}{x}-x^2\right)}{\lim _{x\to \infty \:}\left(8+\frac{2}{x}\right)}=\frac{\lim _{x\to \infty \:}\left(\frac{5}{x}\right)-\lim _{x\to \infty \:}\left(x^2\right)}{\lim _{x\to \infty \:}\left(8\right)+\lim _{x\to \infty \:}\left(\frac{2}{x}\right)}=\frac{0-\infty }{8+0}=\frac{-\infty \:}{8}$

Applying the infinity property: $ \frac{-\infty }{-c}=\infty $

$ =-\infty $

Can someone explain to me why I can’t factor x out?

Let $(X , \cal{A}, m)$ be a measure space. Let $f:X \to [0,1]$ be measurable. If $m(X) < \infty$, find$\lim_{n\to\infty} \int f^n \, d m$.

Let $ (X , \cal{A}, m)$ be a measure space. Let $ f:X \to [0,1]$ be a measurable function. If $ m(X) < \infty$ , determine $ \lim_{n\to\infty} \int f^n \, d m$ .

So far I have:

If $ f(x) < 1$ , then $ \lim_{n\to \infty}{f^n(x)} = 0$ . If $ f(x) = 1$ , then $ \lim_{n\to\infty}{f^n(x)} =1$ . So, for each $ x \in X$ , $ $ \lim_{n\to\infty}{f^n(x)} = \chi_{_{[f = 1]}}(x). $ $

However, I am stuck because I cannot use the Lebesgue Monotone Convergence Theorem, since the sequence is decreasing. Also, I do not know where I will use the hypothesis that $ X$ is a finite measure space. Any ideas?

Let $a_n $ complex sequence prove that if $ a_n\to \infty$ then $|a_n|\to\infty$. Note that $a_n = x_n + y_ni$

Let $ a_n $ complex sequence prove that if $ a_n\to \infty$ then $ |a_n|\to\infty$ . Note that $ a_n = x_n + y_ni$ i dont know how to write that mathmatically.

trial :

Can i say that for every $ M>0$ there exist $ N$ such that for every $ n>N$ ,

$ ~~|x_n|>M~~ OR ~~~|y_n|>M$ ( At least one of them goes to $ \infty$ )

because of that $ |an| = \sqrt{(x_n)^2+(y_n)^2} > M$ and so $ |a_n|\to\infty$ .

$\underset{n\rightarrow +\infty }{\overset{}{\lim }} \ \left(\sqrt[n]{2} -1\right)^{n} =0$

Prove that:

$ $ \underset{n\rightarrow +\infty }{\overset{}{\lim }} \ \left(\sqrt[n]{2} -1\right)^{n} =0$ $

I would like a solution without integral, limit of real functions or others advanced methods. I thought $ \underset{n\rightarrow +\infty }{\overset{}{\lim }} \ 2\left(1- \frac{1}{\sqrt[n]2}\right)^{n} =0$ but I don’t know how to continue.

$\underset{n\rightarrow +\infty }{\overset{}{\lim }} \ \frac{10^{\sqrt{(\ln n)^{2} +\ln( n^{2}})}}{n^{2} +1} =+\infty$

Prove that:

$ $ \underset{n\rightarrow +\infty }{\overset{}{\lim }} \ \frac{10^{\sqrt{(\ln n)^{2} +\ln( n^{2}})}}{n^{2} +1} =+\infty$ $

It is an exercise on first chapters of calculus textbook. I think it is possible to solve without integral or others advanced methods.

Smash product and the integers in a Grothendieck $(\infty, 1)$-topos

Let $ \mathcal{H}$ be a Grothendieck $ (\infty,1)$ -topos. According to this page in nlab, for any $ X \in \mathcal{H}$ , the suspension object $ \Sigma X$ is homotopy equivalent to the smash product $ B \mathbb{Z} \wedge X$ , where $ B \mathbb{Z}$ is the “classifying space of the discrete group of integers.” Furthermore, for any pointed object $ X \in \mathcal{H}_*$ and any group object $ G \in Grp(\mathcal{H})$ , the article says we can “form the tensor product $ X \otimes G \in Grp(\mathcal{H})$ .”

My problem is: none of this terminology is explained, nor does the page provide a reference. Specifically, what is $ \mathbb{Z}$ in an arbitrary $ \infty$ -topos? What is the smash product $ \wedge$ ? What is the tensor product $ \otimes$ ? My best guess is that $ \otimes$ refers to the unique tensor structure on $ \mathcal{H}_*$ such that the map $ \mathcal{H} \to \mathcal{H}_*$ is symmetric monoidal (here $ \mathcal{H}$ is given the Cartesian monoidal structure), but this is only a guess.

Is there a reference where all these notions are defined?