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0

Use geometric series. The limit of the geometric series with ratio $q$ is $\frac{1}{1 - q}$. Compare this with $\frac{1}{2 - x}$ which will lead you to $q = x - 1$. Then one has $$f(x) = \frac{1}{2 - x} = \frac{1}{1 - q} = \sum_{k = 0}^\infty q^k = \sum_{k = 0}^\infty (x - 1)^k.$$ The Taylor series is centered at $x_0 = 1$ and its radius of convergence is ...

1

An easy proof of the quadratic convergence can be found on p. 13-14 of this preprint.

3

Yes, it converges uniformly. No, the M-test will not work - if you could find those $C_n$ that would show the series converges absolutely, which is not so. But: Define $$s_n(x)=\sum_{j=1}^n\frac{\cos(j+x)}{j},$$ $$t_n=\sum_{j=1}^n\frac{\cos(j)}{j},$$ $$r_n=\sum_{j=1}^n\frac{\sin(j)}{j}.$$Then $$s_n(x)-s_m(x)=\cos(x)(t_n-t_m)-\sin(x)(r_n-r_m),$$so $$|s_n(x)-... 0 Here is as elementary a proof as I can come up with that the limit exists and the limit is \sqrt[3]{x_1^2 x_0} . x_{n+2} = \sqrt{x_{n+1} x_n} , so, taking logs, \begin{array}\\ \log x_{n+2} &= \log\sqrt{x_{n+1} x_n}\\ &= \frac12 \log(x_{n+1} x_n)\\ &= \frac12 (\log x_{n+1} +\log x_n)\\ &= \frac12 \log x_{n+1} +\frac12 \log x_n\\ \end{... 3 The existence of such a sequence is impossible. In fact, if \{x_n\} is a sequence of real numbers diverging to \infty, then x_n has a subsequence which, for almost every r, is uniformly distributed mod r. To see this, pass to a subsequence \{x_n'\} of \{x_n\} to get |x_n'-x_m'|>1 for all n\neq m and apply the following Theorem and ... 2 The inequality arises due to a much simpler reason:$$ |x_{in} - x^\ast_{i}|^2 = \color{red}{(x_{in} - x^\ast_{i})^2} \leq \color{blue}{(x_{1n} - x^\ast_{1})^2} + \cdots + \color{red}{(x_{in} - x^\ast_{i})^2} + \cdots + \color{blue}{(x_{mn} - x^\ast_{m})^2} $$The LHS is less than the RHS simply because the RHS includes the LHS \color{red}{\text{red }(x_{in}... 1 Regarding the inequality$$|x_{in} - x_i^*| \le \sqrt{(x_{1n} - x_1^*)^2 + (x_{mn} - x^*_m)^2},$$try squaring both sides. (No triangle inequality needed.) 0 x_{n+2} = \sqrt{x_{n+1} x_n} . Suppose x_n =x_0^{a(n)}x_1^{b(n)}  with a(0) = 1, b(0) = 0, a(1) = 0, b(1) = 1 . Then x_0^{a(n+2)}x_1^{b(n+2)} =\sqrt{x_0^{a(n+1)}x_1^{b(n+1)}x_0^{a(n)}x_1^{b(n)}} =x_0^{(a(n+1)+a(n))/2}x_1^{(b(n+1)+b(n))/2}  so that a(n+2) =(a(n+1)+a(n))/2, b(n+2) =(b(n+1)+b(n))/2 . Both a(n) and b(n) are of the form ru^n+sv^... 1 I think you can say this is true. Lets say that cauchy sequence \{a_i\} in M is related to cauchy sequence \{b_i\} in M if d_M(a_i,b_i)\to 0. We can prove that is an equivalence relation (I think). So if N= the set of all equivalence classes of cauchy sequences then if we consider each m \in M as equal to the equivalence class of all cauchy ... 2 Yes it is true. For a given \epsilon>0 you can find a k such that for n\ge k we have$$0\le x_k\le \epsilon$$Thus the sequence x_n converges to 0 1 Your version is quite long. Here is a hint for a shorter one. What about y_n=\log{x_n} then$$x_{n+2}=\sqrt{x_{n+1}x_n} \Rightarrow \\ \log{x_{n+2}}=\frac{\log{x_{n+1}}+\log{x_n}}{2} \Rightarrow \\ 2y_{n+2}=y_{n+1}+y_{n}$$which is a linear homegenious recurrence which can be solved using characteristic polynomials. I.e. characteristic polynomial is$$2x^2-...

2

Hint Let $b_k=\max_i b_i$, then$$a_n=b_k\sqrt[n]{\left({b_1\over b_k}\right)^n+\cdots +\left({b_m\over b_k}\right)^n}$$Now prove that $$\sqrt[n]{\left({b_1\over b_k}\right)^n+\cdots +\left({b_m\over b_k}\right)^n}\to 1$$by using $$1\le \sqrt[n]{\left({b_1\over b_k}\right)^n+\cdots +\left({b_m\over b_k}\right)^n}\le\sqrt[n]{m}$$and showing (using the ...

1

Notice that $\dfrac{(n+1)(n+1)^n}{(n+2)^n(n+2)}=\left(\dfrac{n+1}{n+2}\right)^{n+1}=\left(1-\dfrac{1}{n+2}\right)^{n+1}$. Recall that for any $y\in\mathbb R$, $$\lim_{n\to +\infty}\left(1+\dfrac{y}{n}\right)^n=e^y.$$

3

Such a sequence exists if you restrict the condition to rational $r>0$. Indeed, you can choose an enumeration $\{q_n\}_{n\geq 1}$ of the positive rationals and for each $n$, we can find an integer $d_n\in \mathbb N$ such that $d_nq_m\in \mathbb N$ for all $1\leq m\leq n$. Then set $$x_n=\prod_{m=1}^n (2d_nq_m),$$ which has the property that $x_n\in q_m\... 2 Of course you can put $$g_n(x)=f_n(x)-f(x)$$ and reduce to the case $$g(x)=0$$ So your book treats the case$f(x)=c$for the sake of simplicity knowing that it is as interesting as the general case 2 No, it does not mean anything. In fact, we define the harmonic numbers as follows: $$H_n=\sum_{k=1}^{n} \frac{1}{k}$$ Because it does not have a nice closed form. But some of them have, for example: $$\sum_{i=1}^{n} i=\frac{n(n+1)}{2}$$ Or $$\sum_{i=1}^{n} i^2=\frac{n(n+1)(2n+1)}{6}$$ 0 There is an asymptotic estimate for this sum, usually denoted by$H_n$, namely $$H_n=\ln n+\gamma+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}+O\left(\frac1{n^6}\right)$$ Up to the Landau big Oh, this is an explicit formula. 1$\frac {a_{n+1}} {a_n}=(1+\frac1 n)^{n} x$so the ratio test tells you that the series converges for$|x| <\frac 1 e$and diverges for$|x|>\frac 1 e$. When$x =\frac 1 e$use Stirling's formula to compare the series with$\frac 1 {\sqrt {2\pi}} \frac 1 {\sqrt n}$to see that the series diverges. 3 You can prove by induction that$0<x_n< 1$for all$n \in \mathbb N$. The derivative of the function$f(x)=\frac{x^3+2}{7}$is$f^\prime(x) = \frac{3}{7}x^2$. Hence$0\le f^\prime(x)\le \frac{1}{2}$for$0 \le x \le 1$. Using the mean value theorem, you get $$\vert f(x)-f(y) \vert \le \frac{1}{2}\vert x -y \vert$$ for$x,y \in [0,1]$. Which leads ... 4 First, show by induction that, with$x_1=1/2$, it follows that$0<x_n\leq 1/2$. Then, since$(x^3 - y^3)=(x^2+xy+y^2)(x-y)$, we find that $$|x_{n+1}-x_{n}| = \frac{|x_n^3 - x_{n-1}^3|}{7}\leq\frac{\frac{1}{2^2} + \frac{1}{2^2}+\frac{1}{2^2}}{7} |x_{n}-x_{n-1}|=\frac{3}{28}|x_{n}-x_{n-1}|.$$ Now use the triangle inequality to estimate$|x_n-x_m|$. Can you ... 2 By induction we see that$0<x_n<1$for all$n$. Hence$|x_{m+1}-x_m| \leq \frac 3 7 |x_m-x_{m-1}|$(by MVT theorem for$x \to x^{3}$). This gives$|x_{m+1}-x_m| \leq (\frac 3 7)^{m-1} |x_2-x_1|$. This implies that$\sum |x_{m+1}-x_m| <\infty$. Can you take it from here? 1 Instead of explicitly demonstrating some natural$K$making the Cauchy condition satisfied, the proof is indirectly doing that by using a known fact from Real Analysis that the sequence$\{C^{n - 1}\}_{n \in \Bbb Z_+}$tends to$0$as$\Bbb Z_+ \ni n \to \infty$as long as$0 < C < 1$(which indeed holds for contractive sequences). In other words, the ... 1 Take$\varepsilon>0$. Take$K\in\mathbb N$such that$$n\geqslant K\implies C^{n-1}\left(\frac1{1-C}\right)\lvert x_2-x_1\rvert<\varepsilon.$$Then, if$m,n\geqslant K$, you have$3$possibilities:$m=n$: then$\lvert x_m-x_n\rvert=0<\varepsilon$.$m>n$: then$\lvert x_m-x_n\rvert\leqslant C^{n-1}\left(\frac1{1-C}\right)\lvert x_2-x_1\rvert<\...

0

An other answer: write $$\frac{w_{n+1}}{w_n}\le \frac{a_{n+1}}{a_n}$$ Fix M>0. The sequence $(\frac{w_n}{a_n})_n$ is descreasing so there is a integer N such that $\frac{w_n}{a_n}< M$ for all $n>M$. Then for all $n>M$, $w_{n+1}\le \frac{w_{n}}{a_{n}}a_{n+1}\le M a_{n+1}$ So $\sum w_n$ is finite. Like $\sum u_{2n+1}$ is finite, hence $\sum u_n$ ...

0

From $$4x_{n+1}=\frac{x_n^2}{4}+\frac{3}{4}$$ we get $$x_{n+1}-x_n=\frac{1}{4}(x_n^2-4x_n+3)=\frac{1}{4}(x_n-2)^2-\frac{1}{4}$$

1

We first note that if the sequence converges to a limit $L\in\mathbb{R}$ then $L$ satisfies the equation $4L=L^2+3$ and we may conclude that $L\in\{1,3\}$. Now let $f(x):=\frac{x^2+3}{4}$, so that $x_{n+1}=f(x_n)$, and in order to discuss the monotonicity of the sequence we consider a few cases. 1) If $x_1=1$ or $x_1=3$ then the sequence is constant (and ...

4

Put $\Bbb T=\{z\in\Bbb C: |z|=1\}$. Suppose to the contrary that for each point $z\in\Bbb T$ the sequence $\{f_n(z)\}$ converges to $f(z)$. Then $f(z)=0$, because, since the set $\{e^{in}:n\in\Bbb N\}$ is dense in $\Bbb T$, for any natural $N$ there exists $n>N$ such that $|z-e^{in}|>1$, so $|f_n(z)|<1/n^2$. Now in order to answer the question we ...

5

Alternative approach If you represent the iterates $x_k$ as fractions $p_k/q_k$, then you have some freedom in assigning the iteration of numerator and denominator in $$\frac{p_{k+1}}{q_{k+1}}=\frac{aq_k}{p_k+q_k}.$$ Choose to make the separate iterations linear, $$\pmatrix{p_{k+1}\\q_{k+1}} = \pmatrix{0&a\\1&1}\pmatrix{p_k\\q_k}.$$ The ...

3

No, because a series of continuous functions cannot converge uniformly to a discontinuous function.

1

It is easy to verify that $|max \{a_1,a_2,...,a_k) - max \{b_1,b_2,...,b_k)|>\epsilon$ implies $|a_i-b_i| >\epsilon$ for some $i$. [Use proof by contradiction]. Hence the result follows immediately from definition of convergence in probability.

1

It is indeed just the ($k$-dimensional) continuous mapping theorem, since $F(x_1, \dots, x_k) := \max(x_1, \dots, x_k)$ is a continuous function from $\mathbb{R}^k$ to $\mathbb{R}$ (easy exercise). You have a sequence of $\mathbb{R}^k$-valued random vectors $\mathbf{X}^n := (X_1^n, \dots, X_k^n)$ converging in probability to a constant vector $\mathbf{x} = (... 1 I give a counterexample: Let us consider the Lebesgue measure$\mu$on$[0,1]$. This is a nice simple probability space. Let$X=0$and $$X_n=\mathbb1_{[(n-2^m)1/2^{m},(n-2^m)1/2^{m}+1/2^{m}]},$$ where$m$is the greatest integer such that$2^m\leq n$. Thus$X_1=1$,$X_2=\mathbb 1_{[0,1/2]}$,$X_3=\mathbb 1_{[1/2,1]}$,$X_4=\mathbb 1_{[0,1/4]}$, etc. Clearly ... 4 Think of what the expressions mean in words. The event$\{|X_n - X| \le \epsilon\}$is demanding that$X_n$is close to$X$.$P(|X_n - X| \le \epsilon) = 1 - o(1)$means that for large$n$, the individual$X_n$must each be close to$X$with high probability. But, the event$\{\sup_{m \ge n} |X_m - X| \le \epsilon\}$is very different - it is demanding ... 0 I consider the$f_{i,k}=\vec{f}_i$and$\lambda_{i,k}=\vec{\lambda}_i$to be sequences in$\Bbb R^K$, and the limit point$f$is a point in$\Bbb R^K$. In your case$K=4$. It would be ideal if the$\vec{\lambda}_i$converged to a limit vector$\vec{\mu} = (\mu_1,\mu_2,\dots,\mu_K)$which satisfied the following system: $$\begin{cases} \sum_{k\geq1} \mu_{k}... 0 Observe that n - (n^2 + 2n)^{\frac{1}{2}}=n - n\left( 1 + \frac{2}{n} \right)^{\frac{1}{2}}. By the binomial theorem, it follows that$$ \left( 1 + \frac{2}{n} \right)^{\frac{1}{2}}=1+\frac{1}{n}-\frac{1}{2n^2}+o(n^{-2}) $$So$$ n - n\left( 1 + \frac{2}{n} \right)^{\frac{1}{2}}=n - \left(n + 1 - \frac{1}{2n} + o(n^{-1}) \right)=-1+\frac{1}{2n}+o(n^{-1})... 0 Let$T$be a tempered distribution. For a Schwartz function$\varphi$then$T \ast \varphi \in C^\infty$. If for some$\varphi$Schwartz satisfying$\int_{-\infty}^\infty \varphi(x)dx = 1$we have$T \ast \varphi \in L^1$then let $$\langle T,1 \rangle = C, \qquad\qquad C = \int_{-\infty}^\infty T\ast \varphi(x)dx$$$\langle T,1 \rangle = C$makes sense ... 0 Set$x=\dfrac 1n\;(x>0)$. You want the limit, as$x\to 0$, of $$\frac1x-\sqrt{\frac 1{x^2}+\frac 2x}=\frac{1-\sqrt{1+2x}}{x}=-\frac{\sqrt{1+2x}-\sqrt 1}{x}$$ This is the opposite of the rate of variation, from$0$, of the function$\sqrt{1+2x}$. Can you end the computation? 3 Hint: Multiply numerator and denominator by $$n+\sqrt{n^2+2n}$$ 3 Note Of course an answer to the question is very simple. Lest anyone get the wrong idea about that from the slightly hairy nature of this post, I point out that there are three parts below: a one-line answer to the question, then an easy generalization using a big theorem, and finally an elementary proof of the big theorem: Answer to the Question Say$...

3

Hint: Write $$\frac{\sqrt{n+1}-\sqrt{n}}{n}=\frac{1}{n(\sqrt{n+1}+\sqrt{n})}$$

12

Can anybody give some hints? Hint $$\frac{\sqrt{n+1}-\sqrt{n}}{n}\color{blue}{\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}}=\frac{1}{n\left(\sqrt{n+1}+\sqrt{n}\right)} \le \frac{1}{2n\sqrt{n}}$$

1

Let $Y_n = \displaystyle \sqrt n\left( \frac 1n \sum_{i=1}^n X_i-\mu\right)$. By the central limit theorem $Y_n$ converges in distribution to some $Y\sim \mathcal N(0,\sigma^2)$, so by the continuous mapping theorem, $|Y_n|$ converges in distribution to $|Y|$. Since $\displaystyle E\left( |Y_n|^2\right) = E\left( Y_n^2\right) = \sigma^2$, $|Y_n|$ is ...

1

You can show that since $\frac{a_n}{a_n+1}$ goes to zero, $a_n$ also goes to zero. Hence it is bounded by, say, $A\in \mathbb{R^+}$. Then you can use the comparison test on (since $a_n\le A$) $$\frac{a_n}{A+1}\le\frac{a_n}{a_n+1}$$ to show that $\sum_{n=0}^\infty a_n$ converges.

1

In case I'm not the only one who thought this was just the definition of unconditional convergence: It appears that here we're saying a series converges unconditionally if any permutation converges, although perhaps to a different sum. Of course the result for finite-dimensional $V$ is immediate from the result for $V=\Bbb R$: Wlog $V=\Bbb R^d$; apply the ...

1

For each $\delta>0$ you have $P(|X_n-X|<\delta/2)\to1$ and also $P(|X_n-\overline X|<\delta/2)\to 1$. So $P(\max(|X_n-X|,|X_n-\overline X|)<\delta/2)\to1$. But the event that $[\max(|X_n-X|,|X_n-\overline X|)<\delta/2]$ is a subset of the event that $[|X-\overline X|<\delta]$. That is, $1\le P(|X-\overline X|<\delta)$, for each ...

2

I'm not familiar with random probability measures... so let me know if I misunderstood something. By assumption (i), we have $$X_n^{(M)} := \int g 1_{|g| \leq M} \, d\mu_n \to X^{(M)} := \int g 1_{|g| \leq M} \, d\mu \tag{1}$$ in distribution for each fixed $M>0$. Consequently, it remains to discuss away the truncation. Set $$X_n := \int g \, d\mu_n \... 1 If you're looking for an elementary proof, one thing you can do is use the identity \arctan(a) = {\pi \over 2} - \arctan(1/a). So for the L^2 limit you're looking at$$\lim_{n \rightarrow \infty} \left(2\int_{0}^{1}\left(\arctan\bigg(\frac{1}{nx}\bigg)\right)^{2}dx\right)^{1/2}$$Changing variables to y = nx this becomes$$\lim_{n \rightarrow \infty} \...

4

Take $f_n(x)=\sqrt n$ for $0 \leq x \leq \frac 1 n$ and $0$ for $x >\frac 1 n$.

1

Yes, it does. From the definition of the distance function, for every natural number $n$ you can find $x_n\in K$ and $y_n\in F$ such that $d(x_n,y_n)<\frac{1}{n}$. Since $F$ is compact, the sequence $y_n$ has a convergent subsequence, say, $y_{n_k}\to y_0$ in $F$. Since $d(x_{n_k},y_{n_k})<\frac{1}{n_k}$, we immediately deduce that $x_{n_k}$ also ...

0

is it $\log \frac{n}{n+1}$? if yes then the first method you did something wrong $\log \left(\frac{n}{n+1} \right) = \log (n) - \log (n+1)$ you cant simplify further $\log(n+1) \ne \log(n) \log (1)$ and in the second method you used the divergence test which just tells you if the limit is not zero then the series is diverge but if it is zero you need to ...

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