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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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