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## Hot answers tagged convergence

8

Show that $$a_2\le a_4\le \cdots\le a_{2n}\le a_{2n-1}\cdots\le a_3\le a_1,$$ i.e., show that $\{a_{2n}\}$ is increasing, while $\{a_{2n-1}\}$ is decreasing. And also $a_{2n}\le a_{2n-1}$. This can be done inductively: $$a_{2k} \le a_{2k+2}\,\,\Rightarrow\,\, \sqrt{2-a_{2k}} \ge \sqrt{2-a_{2k+2}} \,\,\Rightarrow\,\, \sqrt{2-\sqrt{2-a_{2k}}} \le ... 5 Hint: near the fixed point x = 1 the function x\to\sqrt{2-x} is contractive. 4 This iteration x_{n+1} = f(x_n) with f(x) = \sqrt{2-x} has a nice attractive fixed point at (1,1) You can fiddle with the starting value here: GeoGebra interactive worksheet. We have$$ f'(x) = -\frac{1}{2\sqrt{2-x}} $$and$$ \lvert f'(1) \rvert = 1/2 < 1 $$so x^* = 1 is attractive in a neighbourhood. 4 The method is to make a substitution U(n)=\frac{T(n)}{T(n+1)} and you would get much more tangible$$T(n+1)^2-2T(n)T(n+1)-T(n)^2=0$$This one you solve assuming T(n)=a^n and when you substitute and solve you have that a_1=1-\sqrt{2} and a_2=1+\sqrt{2} which gives$$T(n)=c(1-\sqrt{2})^n+d(1+\sqrt{2})^n$$and the solution follows. Set initial ... 4 Outline: Indeed, the key is use Dirichlet's test (a.k.a. Abel's summation at its core) as you intended:$$\begin{align} \sum_{n=1}^N a_n b_n &= \sum_{n=1}^N (A_n-A_{n-1}) b_n = A_Nb_N + \sum_{n=1}^{N-1} A_n b_n -\sum_{n=1}^{N-1} A_n b_{n+1} \\ &= A_Nb_N + \sum_{n=1}^{N-1} \underbrace{A_n}_{\text{bounded}} \underbrace{(b_n -b_{n+1})}_{\text{constant ...

4

You can "weaken" it that way (it turns out to not be a weakening at all, except in apparence). Even take $N=1$ (or any fixed positive number) if you want. Clearly, the first definition implies the second. Now, for the converse: assume we have $\exists N>0 \text{ s.t. }\forall \varepsilon \in (0, N), \exists \delta > 0 s.t. |f(x) - L| < \varepsilon ... 3 Hints: it converges normally on any$I_\delta \stackrel{\rm def}{=}(-\infty, \delta)\cup(\delta, \infty)$(for any fixed$\delta > 0$. Indeed, for all$n\geq 0$the function$f_n$is even, non-negative, and decreasing on$(\delta,\infty)$, so that $$\sup_{x\in I_\delta} \lvert f_n(x)\rvert = \sup_{x\in I_\delta} f_n(x) = \frac{1}{1+n^a\delta^4}.$$ ... 3 For$a>1$and$x\ge \delta>0$, we have $$\sum_{n=1}^\infty\frac{1}{1+n^ax^4}\le \sum_{n=1}^\infty\frac{1}{n^a\delta^4}=\frac{1}{\delta^4}\zeta(a)$$ which exists for$a>1$, which one can show using, say, the integral test. However, we can choose a number$\epsilon=\frac12$, and a number$x=1/n^{a/4}$such that for any$n$... 3 You have surely proved that$a_n\le 2$for all$n$. Consider the sequences$b_n=a_{2n-1}$and$c_n=a_{2n}$. The recursions are $$b_{n+1}=a_{2n+1}=\sqrt{2-a_{2n}}=\sqrt{2-\sqrt{2-a_{2n-1}}}= \sqrt{2-\sqrt{2-b_n}}$$ Let's show$(b_n)$is decreasing: \begin{gather} b_{n+1}\le b_n\\ \sqrt{2-\sqrt{2-b_n}}\le b_n\\ 2-\sqrt{2-b_n}\le b_n^2\\ 2-b_n^2\le ... 2 Convergence speed: For$x=1+t$close to$1$, $$1+t_{n+1}=\sqrt{1-t_n}\approx1-\frac{t_n}2$$ and $$t_{n+k}\approx t_n\left(-\frac12\right)^k.$$ The convergence is linear (one more exact bit per iteration). This is well illustrated by a logarithmic plot of the residue: 2 Hint: write$a_n=1+b_n$or$a_n=1-b_n$, whichever makes$b_n$positive. How does$b_n$behave? Elaboration: we have$a_n=1+b_n$for odd$n$and$a_n=1-b_n$for even$n$(why so?). So, for example, for even$n$we can write$a_{n+1}=\sqrt{2-a_n}$as$1+b_{n+1}=\sqrt{2-(1-b_n)}=\sqrt{1+b_n}$. Now you can compare$b_{n+1}$and$b_n$. Proceed similarly for odd ... 2 This is correct, but to be completely legit (or even just pass the test of "proof-checking"), you should add the following details: boundedness: give the details of the induction, if you want us to check it; monotonicity: the sequence is indeed increasing, but how did you show it? (e.g., "the function$x\mapsto\sqrt{3x-2}-x$is positive on$[3/2,2)$") ... 2 Is easy to check that for$x>\sqrt2-1$,$x>1/(2+x)>\sqrt2-1$, so $$U_{n-1} >\sqrt2-1\implies \sqrt2-1<U_n = \frac1{2+U_{n-1}} < U_{n-1}.$$ Starting from$U_1>\sqrt2-1$, this proves that the sequence is decreasing and bounded, so convergent. Now, take limits in$U_n = \frac1{2+U_{n-1}}$. 2 If there are$N\in{\mathbb N}$and$m\in{\mathbb Z}$as described then the sequence$(a_n)_{n\geq0}$is obviously convergent. Conversely, if the sequence$(a_n)_{n\geq0}$is convergent it is a Cauchy sequence. So there is an$N\in{\mathbb N}$with$|a_n-a_N|<1$for all$n>N$. As all$a_n$are integers this implies that in fact$a_n=a_N=:m$for all ... 2 It depends how we "filter out." If we remove all the$b_n$we may end up with a sequence that is finite, or even empty. But if we remove say$b_1$and$b_4$and$b_9$and$b_{16}$and so on, then we are left with an infinite sequence and can repeat the process. 2 We have$x_n \to 0$. Hence there is some$N$such that$n \geq N$implies$x_n > -1/2$. Therefore when$n \geq N$, we have $$\left| \frac{x_n}{1 + x_n}\right| = \frac{|x_n|}{1 + x_n} \leq 2|x_n|.$$ Thus the series$\sum \frac{x_n}{1 + x_n}$converges absolutely by comparison with$\sum |x_n|$. 2 Here is a hint, since I don't know what your thoughts on the problem are. Pick an arbitrarily large$M>0$. Can you show that $$\frac{2^n}{n}\geq M$$ for every big enough$n$? 2 Hint Use the betweeness property to pick some$a_n$between$x$and$x-\frac{1}{n}$. 2 Hint: Assume$(x_n)$is C-convergent to$x$. Fix$n > N$. Then for all$\epsilon > 0$we know that$0 \leq d(x_n,x) < \epsilon$. So$d(x_n,x) \in \cap_{\epsilon > 0} [0,\epsilon) = \{0\}$, which is to say that$x_n = x$. Thus a sequence$(x_n)$is C-convergent if and only if there exists$x \in \mathbb{R}$and$N$such that$x_n = x$for all$n ...

2

The definitions are equivalent. Assume that there exists $N > 0$ such that for any $\varepsilon \in (0, N)$ there exists $\delta > 0$ for which $0 < |x - a| < \delta$ implies that $|f(x) - L| < \varepsilon$. We want to show that given $\varepsilon' > 0$ there exists $\delta' > 0$ for which $0 < |x - a| < \delta'$ implies that ...

2

We have to assume $\mathbb{E}(|X|^r)<\infty$; otherwise the expession $\|X_n-X\|_{L^r}$ might not even be finite. Note that $$\|X_n-X\|_{L^r}^r = \int_{|X_n-X| \leq \epsilon} |X_n-X|^r \, d\mathbb{P} + \int_{|X_n-X| > \epsilon} |X_n-X|^r \, d\mathbb{P} \tag{1}$$ for any $\epsilon>0$ and $n \in \mathbb{N}$. Obviously, $$\int_{|X_n-X| \leq ... 2$$ a_{n+1}^2-a_n^2=6+a_{n}-a_n^2=(3-a_{n})(2+a_n) If a_n>3, a_{n+1}>\sqrt{6+3}=3. So by induction a_n>3\;\forall\;n and a_{n+1}^2<a_{n}^2\;\forall\;n. And only possible limit is the positive solution of x^2=6+x. 2 Note that the X_n are bounded by 1, so the absolute convergence follows from the convergence of the sum \sum \limits_{n = 1}^\infty \frac{2}{3^n}. 1 This is my try. Let: f_n (y) = \frac{e^y}{n^2y^4+1} \mathbb I_{[0,1]} , where \mathbb I is indicator function. We have: f_n(y) \to 0, as n \to \infty. For integrability and domination condition, for every n \in \mathbb N, |f_n(y)| \leq \frac{e^y}{y^4+1} \mathbb I_{[0,1]} \leq e^y \mathbb I_{[0,1]}. This function is integrable on \mathbb R, ... 1 Formally, there is a probability space Ω (with a σ-algebra \mathcal F and a probability measure P) and the random variables X_n map Ω to \mathbb R (with \mathcal B(\mathbb R)) as follows \begin{align}X_n:Ω&\mapsto \mathbb R\\[0.2cm]ω &\to \{0,1\} \end{align} with P(\{ω:X_n(ω)=1\})=P(\{ω:X_n(ω)=1\})=1/2. Now, take an ω \in Ω and ... 1 First we prove by induction that a_n>3. It's true for n=1. Assuming a_n>3, we know a_n+6>3^2 so \sqrt{a_n+6}>3 or a_{n+1}>3. Thus we established the lower bound 3. Now we see that x^2-x-6 is a strictly increasing polynomial for x>3, and has a root at x=3, thus, x^2-x-6>0 for x>3: We see that ... 1 Your intuition is pretty good. You actually claimed that it doesn't matter what the function does "far away" from f(x_0) (More precisely, outside some neighborhood of f(x_0)). 1 Looks like you're applying the ratio test to determine when the series converges. There are a couple of typos in your formulas and your typesetting. Here's how the reasoning should look: Write u_n=(2n-1)x^n. Then {u_{n+1}\over u_n} = {(2(n+1)-1 ) x^{n+1}\over (2n-1)x^n} = x\cdot {2n+1\over 2n-1} = x\cdot {2+\frac1n\over 2-\frac 1n}, $$so$$\lim ...

1

$C$-convergence is much stronger than convergence. The sentence for every $\epsilon>0$ and every $n>N$, $d(x_n,x)<\epsilon$ is the same as for every $n>N$, $x_n=x$.

1

For all $n \in \mathbb{N}$ there is by the "Betweenness Property" a rational number $x + \frac{1}{n} > a_n > x$. Can you do the rest?

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