5
votes
How to solve this mathematical analysis limit problem?
Let $a_n$ denote $\sum_{k=1}^{n}\left(\dfrac{1}{n}+\dfrac{k}{n^2}\right)^{1+\frac{2k}{n^2}}$
Since we know $\frac{1}{n}+\frac{k}{n^2}<1$, Then:
$a_n \leq$ $\sum_{k=1}^{n}(\frac{1}{n}+\frac{k}{n^2})...
- 1,142
5
votes
Accepted
Given continuous $f:\mathbb{R}\to\mathbb{R}$ and compact $K\subseteq f(\mathbb{R})$, show there exists a compact set $C$ such that $f(C)=K$.
Let $a=\inf K$ and $b=\sup K$. By Let $f : I \to \mathbb R$ be continuous. For any compact interval $J \subseteq f(I), \exists$ a compact interval $K \subseteq I$ with $f(K)=J.$
we see that there is ...
- 24.5k
4
votes
Accepted
Proving $\lim_{n \to \infty}\int_n^{n+1}\frac{\sqrt{x-1}}{\sqrt{x^2+1}}dx=0$
We have that
$$0\le \int_n^{n+1}\frac{\sqrt{x-1}}{\sqrt{x^2+1}}dx \le \int_n^{n+1}\frac1{\sqrt{x}}dx=2(\sqrt{n+1}-\sqrt n)=\frac2{\sqrt{n+1}+\sqrt n}\to 0 $$
As noticed in the comments, more simply
$$...
- 148k
4
votes
Accepted
Is $\mathbb{R}\cup\{-\infty,+\infty\}$ the categorical co-completion of $\mathbb{Q}$
The extension $L$ you describe does not preserve colimits in general. For instance, let $\mathcal{E}$ be the poset $\{0,1\}$ and let $D$ send the negative rationals to $0$ and the nonnegative ...
- 317k
4
votes
Accepted
Why does $\lim_{x\to\infty}\prod\limits^{\lfloor x\rfloor}_{n=1}\frac{\lceil x\rceil}{\lfloor x\rfloor}=e$?
Let $k=\lfloor x\rfloor$
$$\prod\limits^{\lfloor x\rfloor}_{n=1}\frac{\lceil x\rceil}{\lfloor x\rfloor}=\prod\limits^{k}_{n=1}\frac{k+1}{k}=\left( 1+\frac1k\right)^k$$
therefore,
$$\lim_{x\to\infty}\...
- 18.4k
3
votes
Accepted
Find the function $f(x)=\sum_{n=1}^{\infty}3(-1)^{n+1}(3x)^{n-1}$
$$f(x)=\sum_{n=1}^{\infty}3(-1)^{n+1}(3x)^{n-1}$$
Let $m=n-1,$
$$f(x)=\sum_{m=0}^{\infty}3(-1)^{m+2}(3x)^{m}=3\sum_{m=0}^{\infty}(-3x)^{m}=\frac{3}{1+3x}$$
- 18.4k
3
votes
Proving $\lim_{n \to \infty}\int_n^{n+1}\frac{\sqrt{x-1}}{\sqrt{x^2+1}}dx=0$
Define $$F(x) := \int_1^x \frac{\sqrt{x-1}}{\sqrt{x^2+1}}\, dx$$
By the Fundamental Theorem of Calculus, $F$ is differentiable, and
$$F'(x) = \frac{\sqrt{x-1}}{\sqrt{x^2+1}}$$
for all $x > 1$.
...
- 12.4k
2
votes
Compute the integral $\int_0^\infty t^{3/\xi-1} e^{-t} \Gamma(\frac{2}{\xi},t) \ dt$
The only closed form I see is via the incomplete beta function: $$\int_0^\infty t^{\alpha-1}e^{-t}\,\Gamma(\beta,t)\,dt=\Gamma(\alpha+\beta)\,\mathrm{B}(1/2;\alpha,\beta),$$
obtained using the ...
- 37.4k
2
votes
Accepted
sequence of non-negative functions $f_n$ tending to $0$ pointwise with $\int f_n \to 0$, but no integrable $g$ such that $f_n \leq g$ for all $n$.
One way of coming up with an example is to let $f_n$ be a triangle function supported on $[n,n+1]$ with height $\frac{2}{n}$, so that $\int f_n=\frac{1}{n}$. Then, $f_n\to 0$ pointwise and $\int f_n\...
- 46k
2
votes
Is "uniform continuity theorem" provable without using sequences?
You can, in fact, prove uniform continuity in this style… with a slight modification.
(The flaw in Cactus’s counter-argument – though their proof is the “better” one – is that even for a function $\...
- 31.7k
2
votes
The set $X := [1, 2] \cup [3, 4]$ is not connected
When we are dealing with metric subspaces like $X$, the notion of something being "relatively open" is exactly equivalent to openness in the subspace.
That is, we have that for instance $[1,...
- 283
2
votes
Is $\mathbb{R}\cup\{-\infty,+\infty\}$ the categorical co-completion of $\mathbb{Q}$
The free cocompletion of any category is its category of presheaves. The category of presheaves on $\mathbb Q$ is the category of families of sets, contravariantly indexed by rational numbers. This is ...
- 50.8k
1
vote
Is "uniform continuity theorem" provable without using sequences?
You can't do it like that. Indeed, if such an argument worked, you could use the same reasoning to prove that any function $\mathbb{R} \rightarrow \mathbb{R}$ is uniformly continuous on $\mathbb{R}$ (...
- 1,732
1
vote
Decide whether $f$ is differentiable in $(0,0)$ or not given its directional derivative
If $f$ is differentiable in $a$ then you know that for $v = (v_1,v_2)$ you should have
$$ \frac{\partial f}{\partial v} (a)= (\nabla f (a)) \cdot v = v_1 \frac{\partial f}{\partial e_1}(a) + v_2 \...
- 2,796
1
vote
Accepted
Prove $\exp(\int^x_0 \frac{U(t)}{t} dt)$ is reguarly varying (Extreme Values, Regular Variation and Point Processes)
Let's denote $f(x) = \exp \left(\int\limits_0^x \frac{U(t)}{t} {\rm d}t\right)$. Following the given theorem, one needs to prove that $\int\limits_0^1 \ln \left(\frac{f(x)}{f(\lambda x)}\right) {\rm d}...
- 2,197
1
vote
The set $X := [1, 2] \cup [3, 4]$ is not connected
Openness and closedness depends on the ambient space. In the very common ambient space of the real numbers, $[1,2]$ is a non-open closed subset, as there does not exist any $r > 0$ such that we ...
- 1,034
1
vote
The set $X := [1, 2] \cup [3, 4]$ is not connected
I think you can agree that in any metric space $(X,d)$, the open balls $B_r(x)$ with radius $r$ centered at $x$ are open. In fact, by definition a subset $Y\subseteq X$ of a metric space is open if ...
- 11.8k
1
vote
Can this proof of the second MVT for integrals be salvaged?
It seems unlikely that we can extend this argument to cases where $f$ changes sign on $[a,b]$. We can easily construct examples where $\int_a^b fg$ does not lie between $A$ and $B$.
Let $g(x)=x$, $f(x)...
- 3,159
1
vote
Accepted
A question about the d'Alembert ratio test for convergence.
Suppose we do have some sequence $\{ a_n \}$ that fulfills the property as described, i.e. $\lim_{n \rightarrow \infty}|\frac{a_{n + 1}}{a_n}| = L > 1$. With this assumption, we can choose some $\...
- 283
1
vote
Local minima and derivative
transfer everything except for o(h) on the LHS, use the definition of o(h) - particularly the definition of a limit (epsilon delta definition), put epsilon := (1/2)|f′(x∗)|. Normal people use epsilon ...
- 31
1
vote
Accepted
Can the convergence of inner products be 'closed'?
What follows is an example that the uniform convergence on $K$ does not suffice. Let $\{e_j\}_{j=1}^\infty$ be an orthonormal basis. Consider
$$v_n=\sum_{j=1}^n 2^{-j}e_j,\quad v=\sum_{j=1}^\infty 2^{-...
- 21.3k
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