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2h
comment Sequence of functions that converge uniformly on a compact metric space.
You identified the problem: boundedness. I suspect the intention was to have the functions continuous, for otherwise the compactness of the domain would be utterly irrelevant. However, only $f$ needs to be bounded. If you don't throw away the denominator, the factor of $\lvert f(x) - f_n(x)\rvert$ becomes $\frac{1 + \lvert g(x)\rvert}{(1 + \lvert g_n(x)\rvert)(1 + \lvert g(x)\rvert)}$, which is evidently bounded (by $1$). But for unbounded $f$, you can take $f_n =f$ for all $n$, and simple $g_n$ to see that $\frac{f}{1 + \lvert g_n\rvert}$ need not converge uniformly.
5h
comment Asymptotic behavior of elliptic integral (first kind)
looks promising. From then on, it's pretty straightforward, one has $\sqrt{1+t^2} \cdot \sqrt{\varepsilon^2 + t^2}$ in the denominator, and to simplify the analysis one isolates the singular (as $\varepsilon \to 0$) factor by replacing the well-behaved one with its limit at $t = 0$.
5h
comment Asymptotic behavior of elliptic integral (first kind)
That's the streamlined version. After playing around a little with it, it turned out that $t = \tan \varphi$ gives an integral one can deal with fairly well, then I rewrote the $1$s so that the derivative of $\tan$ appears directly. The thing is that one has to try a couple of substitutions that occur often and see whether one leads to success. It's fairly natural that one tries to pull out a $\sin \varphi$ or a $\cos \varphi$ from $\sqrt{\sin^2\varphi+\varepsilon^2\cos^2 \varphi}$. Then when one sees $\cos \varphi\cdot \sqrt{\varepsilon^2 + \tan^2 \varphi}$, it's clear that $t = \tan\varphi$
22h
comment Is $E$ path connected $\implies \overline{E}$ connected?
But be aware that $\overline{E}$ need not be path-connected.
22h
comment Dirichlet's Test in convergence
@DavidC.Ullrich Answer at the duplicate target, that isn't closed. (And that answer would be sure to get an upvote from me.)
1d
comment Only once differentiable
If you are happy with examples where $f''(x)$ doesn't exist at finitely many points, explicit examples are easy to give. If you want $f''(x)$ to exist at no point, examples are harder to come by.
1d
comment Can Stirling's approximation be used to obtain lower and upper bound for $\pi(x)$?
No. If we denote the approximation of $k!$ by $\alpha(k)$, then you need $\lvert (j-1)! - \alpha(j-1)\rvert < 2j$ to be able to say anything about the sine term (beyond the trivial $0 \leqslant \sin^2 x \leqslant 1$), that's much much more than Stirling's approximation gives you.
1d
comment Endomorphisms of $\mathbb C^{\times}$
As a topological group, $\mathbb{R}_{> 0}^\times$ is isomorphic to $(\mathbb{R},+)$. Does that give you an idea?
1d
revised Do nested integrals exist?
Typo fixes
2d
comment Finite ring having $2^n-1$ invertible elements
@user265311 Since the characteristic is $2$, it is a $\mathbb{F}_2$ vector space.
2d
comment Finite ring having $2^n-1$ invertible elements
@hardmath Well, yes, here we have the bounds $2^n - 1 < \lvert A\rvert \leqslant 2\cdot 2^n - 1$, I figured that doesn't make much of a difference.
2d
comment Are there other known continued fractions that show the digits of the golden ratio?
Those are the continued fraction expansions of $10^k\cdot \frac{1+\sqrt{5}}{2}$?
2d
comment Is the magnitude of the gradient non zero?
Bingo. And by assumption, $f(z_0) \neq 0$.
2d
comment Is the magnitude of the gradient non zero?
Right. So by assumption, it's $> 0$. And it's the determinant of the matrix.
2d
comment Is the magnitude of the gradient non zero?
What is $u_x v_y - u_y v_x$?
2d
comment What is the sum of this series? $\sum_{n=1}^\infty\frac{2^n (-1)^{n+1} B_n}{n}$
Hm? $$\frac{1}{x+\frac{1}{2}} = \frac{2}{1+2x} = 2\sum_{n = 0}^\infty (-1)^n2^nx^n,$$ no Bernoulli numbers to be seen in that Maclaurin expansion.
2d
comment Is the magnitude of the gradient non zero?
Ignoring the constant factor $2$, we can write that as $$\begin{pmatrix} u_x & v_x \\ u_y & v_y\end{pmatrix}\cdot \begin{pmatrix} u\\ v \end{pmatrix}.$$ See something?
2d
comment What is the sum of this series? $\sum_{n=1}^\infty\frac{2^n (-1)^{n+1} B_n}{n}$
What are the $B_n$? Bernoulli numbers? Then the series is divergent. So if you want to assign a value to it, you should probably specify the desired summation method.
2d
comment Is the magnitude of the gradient non zero?
Can you compute $\nabla F$?
2d
comment Non-standard numbers and exponential form of Zeta function
No, it describes some properties which you think your objects have. That is very far from an exact specification.