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3

First of all, we choose $M \in \mathbb{N}$ sufficiently large such that $\sum_{m \geq M} 2^{-m} < \delta/3$. Moreover, we note that \begin{align*} \mathbb{E}\left( 1 \wedge \sup_{s \leq m} |X_s^n-X_s| \right) &\leq \mathbb{E}\left( 1 \wedge \sup_{s \leq m} |X_s^n-X_s| \cdot 1_A \right)+ \mathbb{E}\left( 1 \wedge \sup_{s \leq m} |X_s^n-X_s| \cdot ... 0 Your example is not true, because g_n = \chi_{[n,n+1]} isn't Cauchy measure so this sequence isn't convergence to 0 in measure. 1 If you want to prove that a_n approaches 0 faster than b_n, you usually need to prove that\lim_{n\to\infty}\frac{a_n}{b_n}=0.$$In your case, you want to show that$$\lim_{n\to\infty}\frac{x^n\cdot x^n}{n!} = 0.$$This is no more difficult than proving that \frac{x^n}{n!} has a limit of 0 for any value of x, since$$\frac{x^n\cdot x^n}{n!} = ...

1

Alternatively you can show that $(s_n)$ [the $n$-th partial sum] is a Cauchy sequence: Denote $S_n$ the $n$-th partial sum for the absolute values of $b_n$, i.e., $S_N =\sum_{n=0}^N |b_n|$. Choose $n_0$ such that $S_q-S_p<\varepsilon$, for all $q>p\ge n_0$. Then $$|s_q-s_p|=\bigg|\sum_{n=p+1}^q b_n \bigg|\le\sum_{n=p+1}^q |b_n|=S_q-S_p< ... 1 You know that if we have two convergent sequences (in the wide sense of the word) \;\{a_n\}\;,\;\;\{b_n\}\; s.t. \;a_n\le b_n\; , then \;\lim a_n\le\lim b_n\; . Well, now apply this to the sequences of partial sums by means of the triangle inequality:$$\forall\,N\in\Bbb N\;,\;\;\left|\sum_{n=1}^na_n\right|\le\sum_{n=1}^N|a_n|\implies ...

0

There is an $r\ge a$ such that $$\frac{f(x)}{g(x)}\le 1 \quad \forall x \ge r.$$ It follows that $$0\le \int_a^\infty f(x)\,dx\le \int_a^rf(x)\,dx+\int_r^\infty g(x)\,dx<\infty.$$

1

It may not be wrong that $\lim_{x\to+\infty}f(x)\to0$, but it could be that $g(x)$ grows faster than $f(x)$, and so maybe that's why $\lim_{x\to+\infty}\dfrac{f(x)}{g(x)}\to0$. So, it's not a flaw in logic - it's just that you are removing other plausible situations.

1

Let $S(x)$ your sum. On its domain: $$\frac d{dx}x^3S(x) =\frac d{dx} \sum_{n=0}^\infty \frac{x^{n+3}}{n+3} \\ = \sum_{n=0}^\infty x^{n+2} = \frac {1}{1-x} - 1 - x$$ So the radius is $R= 1$ [it is invariant under derivation and multiplication by $x^3$] and then you integrate and divide by $x^3$.

3

Notice that for $x=1$ the series $\sum_n \frac1{n+3}$ is divergent then the radius of convergence $R\le1$ but for $x=-1$ the series $\sum_n \frac{(-1)^n}{n+3}$ is convergent by Leibniz theorem so $R\ge1$. Conclude. For the sum we have $$\sum_{n=0}^\infty \frac{x^n}{n+3}=\sum_{n=3}^\infty \frac{x^{n-3}}{n}=\frac1{x^3}\sum_{n=3}^\infty ... 1 Let$$u_n=\frac{x^{2n}}{\sqrt{n+1}}$$then by ratio test we have easily$$\lim_{n\to\infty}\left|\frac{u_{n+1}}{u_n}\right|=|x|^2<1\iff|x|<1$$can you take it from here? 1 \begin{eqnarray} S&=&\sum_{i=1}^\infty\frac{F_i}{10^{i+1}}=\frac{F_1}{10^2}+\frac{F_2}{10^3}+\sum_{i=3}^\infty\frac{F_i}{10^{i+1}}=\frac{F_1}{10^2}+\frac{F_2}{10^3}+\frac{1}{10^3}\sum_{i=1}^\infty\frac{F_{i+2}}{10^i}\\ ... 1 You can use the ratio-test to prove the convergence of the series, but that does not prove that the limit is a rational number. But the generating function$$ s(x) = \sum_{n=0}^\infty F_n x^n $$has the closed form s(x) = x/(1-x-x^2), see for example http://en.wikipedia.org/wiki/Fibonacci_number#Power_series. Therefore$$ \sum_{n=1}^\infty ...

1

The series $\displaystyle\sum_{n\ge0} a_n$ is convergent if and only if the partial sum sequence $\left(\displaystyle\sum_{k=0}^n a_k\right)_n$ is convergent which's equivalent to the convergence of the sequence $\left(\displaystyle\sum_{k=n_0}^n a_k\right)_n$ for all $n_0\in\Bbb N$ hence the convergence of a series doesn't depend of its first few terms. ...

1

Here is another way using the Cauchy's criterion: Let $M>0$ such that $|b_n|< M$. Given $\varepsilon>0$, choose $n_0$ such that $$\sum_{n=p+1}^q|a_n|< \varepsilon/M$$ for all $q<p<n_0$ Then $$\bigg|\sum_{n=p+1}^qa_nb_n\bigg|\le\sum_{n=p+1}^q|a_nb_n|\le M\sum_{n=p+1}^q|a_n|< \varepsilon$$

0

Absolute values are critical: consider $a_n=(-1)^n/n$ and $b_n=(-1)^n$. Luckily you are given absolute convergence so use it to show $\sum a_nb_n$ converges absolutely.

1

$\textbf{Hint:}$ Let $\displaystyle S_{k}=\sum_{n=1}^k a_{n}$ and $\displaystyle T_{k}=\sum_{n=1}^k |a_{n}|$ $\;$ for $k\ge1$, and let $\displaystyle S=\sum_{n=1}^{\infty}a_{n}$ and $\displaystyle T=\sum_{n=1}^{\infty}|a_{n}|$; so $S=\displaystyle\lim_{k\to\infty}S_{k}$ and $T=\displaystyle\lim_{k\to\infty}T_{k}$. By the Triangle Inequality, ...

0

Assuming continuity at $x=0$, $n\int\limits_0^1 f(x)e^{-nx}\,dx=\int\limits_0^1 f(x)\,d(e^{-nx})$ where, $\lim\limits_{n\to\infty}e^{-nx}=1 (\text{when, }x=0) \text{ and } 0\text{ when } 1\ge x>0$. So, $\lim\limits_{n\to\infty} n\int\limits_0^1 f(x)e^{-nx}\,dx = f(0)$.

2

Hint: We can say :$$\omega \in C \Leftrightarrow (\forall m \in \Bbb N )(\exists p \in \Bbb N)( \forall k \geq p) \quad \left|\sum_{i=k+1}^{+\infty} X_k(\omega)\right| < \frac 1{m+1}$$ That gives:$$C=\bigcap_{m \in \Bbb N}\bigcup_{p \in \Bbb N}\bigcap_{k \geq p} Y_{m,p,k}$$ Where : $$Y_{m,p,k}=\left\{\omega \in \Omega / \left| \sum_{i=k+1}^{+\infty} ... 0 Hint. Note that if \lvert f(x)\rvert\le M, then$$ \left|\,n\int_0^1 f(x)\,\mathrm{e}^{-nx}\,dx\,\right|\le Mn\int_0^1 \mathrm{e}^{-nx}\,dx= -nM \frac{\mathrm{e}^{-nx}}{n}\Big|_0^1=M-M\mathrm{e}^{-n}\to M. $$In fact if f is continuous at x=0, then$$ \lim_{n\to\infty} n\int_0^1 f(x)\,\mathrm{e}^{-nx}\,dx=f(0). $$1 As @Claude has stated for the simpler cases here is it for 4 elements in a and in b. The correct sum is the sum over all elements of the ("outer"(?)) product C of the two vectors$$ A^T \cdot B=C= \small \begin{array} {r|rrrr} & b_0 & b_1 & b_2 & b_3 \\ \hline a_0 & a_0b_0 & a_0b_1 & a_0b_2 & a_0b_3 \\ a_1 & ...

2

Just try with two terms. So, you have $$(\sum_{i=0}^1 a_i)(\sum_{i=0}^1 b_i)=a_0 b_0+a_1 b_0+a_0 b_1+a_1 b_1$$ $$\sum_{i=0}^1 \sum_{j=0}^1 a_jb_{i-j}=a_0 b_0+a_1 b_0+a_0 b_1$$

3

This is certainly not true: $$\left(\sum_{i=0}^n a_i\right)\left(\sum_{i=0}^nb_i\right) = \sum_{i=0}^n \sum_{j=0}^i a_jb_{i-j}$$ So it's not clear why you'd expect the limit as $n\to\infty$ is the same. If definitely requires a much more careful argument. The actual equality is: $$\left(\sum_{i=0}^n a_i\right)\left(\sum_{i=0}^nb_i\right) = \sum_{i=0}^n ... 1 If you stick to$$a_n={1!+2!+\cdots+n!\over(2n)!}$$then$$\begin{align} \left|{a_{n+1}\over a_n}\right|&={(2n)!\over(2n+2)!}\left({1!+2!+\cdots+n!+(n+1)!\over1!+2!+\cdots+n!}\right)\\ &={1\over2(n+1)(2n+1)}\left(1+{(n+1)!\over1!+2!+\cdots+n!} \right)\\ &={1\over2(2n+1)}\left({1\over n+1}+{n!\over1!+2!+\cdots+n!}\right)\\ ...

1

Let $s_n = \sum_{i=1}^{n} \frac{1}{2^i}$. Note that $s=\sum_{n=1}^{\infty} \frac{1}{2^n} < \infty$. Consider $\tilde{a_n} = a_n + s_{n-1}$. We have that $(\tilde{a}_n)$ is bounded. Also, we have $\tilde{a}_{n+1} = a_{n+1} +s_{n} \ge a_n - \frac{1}{2^n} + s_{n} = a_n + s_{n-1} = \tilde{a}_n$ so that $(\tilde{a_n})$ is increasing. By the monotone ...

7

Since $(2n)!=(n!)^2\dbinom{2n}{n}>(n!)^2$, then $$\sum_{n=1}^{\infty}\frac{1!+\dots+n!}{(2n)!}<\sum_{n=1}^{\infty}\frac{n\cdot n!}{(n!)^2} =\sum_{n=1}^{\infty}\frac{1}{(n-1)!}=e<\infty.$$

0

The empty set is open in this definition, because the condition on the left (the non-empty intersection with $O$) is never fulfilled for the empty set, of course. And an implication whose left clause is false, is a true statement (ex falso totum). The intersection argument is correct, but could be more explicit: suppose that $O_1$ and $O_2$ are open ...

1

It's probably simpler to prove that the complement of $\limsup_{n\to+\infty}E_n$ is contained in $C$. Indeed, if $\omega$ is not in $\limsup_{n\to+\infty}E_n$, then there is $N=N(\omega)$ such that for $n\geqslant N$, $\omega\notin E_n$. We thus have $|X_n(\omega)|\leqslant q^n$ for these $n$, which proves the convergence of the series $\sum_{n\geqslant ... 3 We have $$1-\frac{a_n}{a_{n+1}}=\frac{a_{n+1}-a_n}{a_{n+1}}\le\frac{a_{n+1}-a_n}{a_{1}}$$ and the telescoping series $$\sum_n{a_{n+1}-a_n}$$ is convergent since the sequence$(a_n)$is convergent so the given series is convergent by comparison. 2 The sequence$(a_n)_n$necessarily converges to some$l$, hence the ratio test used in the OP attempt is not conclusive. However, since$0\leqslant \frac{a_{n+1}-a_n}{a_{n+1}}\leqslant \frac{a_{n+1}-a_n}{a_1}$, the$N$-th partial sum is bounded by$\frac{a_{N+1}-a_1}{a_1}$, which is itself bounded by$\sup_N(a_{N+1}-a_N)$. 1 Take in to major account Did's comment. If you properly apply the definition and build your series at$x=0$, you will obtain $$1+x+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^5}{15}-\frac{x^6}{240}+\frac{x^7}{90}+\frac{3 1 x^8}{5760}+O\left(x^9\right)$$ 0 Usually the definition of epi-convergence is made for lsc functions only. Even if this was not the case, the limit would be lsc automatically. So, if a function is not lsc, you cannot hope to epi-approximate it by nicer functions, which is a main purpose of epi-convergence. (It is used to transfer information about the minima of nice functions to their ... 2 For any norm$\|\cdot\|$, by the triangle inequality, $$\|X_n\|\le\|X_n-X\|+\|X\|$$ and $$\|X\|\le\|X-X_n\|+\|X_n\|$$ so that $$|\|X_n\|-\|X\||\le\|X_n-X\|.$$ This property is sometimes called the continuity of the norm.$(\operatorname E|X|^r)^{1/r}$is the norm of the space of random variables with$\operatorname E|X|^r<\infty$for$r\ge1$. We ... 15 We have that $$\left(1+\frac{1}{n}\right)^n=\mathrm{e}^{n\log(1+\frac{1}{n})}=\mathrm{e}^{1-\frac{1}{2n}+{\mathcal O}(n^{-2})}=\mathrm{e}\left(1-\frac{1}{2n}+{\mathcal O}\Big(\frac{1}{n^2}\Big)\right),$$ since $$\log \Big(1+\frac{1}{n}\Big)=\frac{1}{n}-\frac{1}{2n^2}+{\mathcal O}\Big(\frac{1}{n^3}\Big) \quad\text{and}\quad \mathrm{e}^h=1+h+{\mathcal ... 0 You are right that if the Fourier series is absolutely convergent, then you can do this sort of rearrangement. But in general you should not expect the coefficients of Fourier series to be absolutely convergent. (Nonzero functions for which this does hold are members of what is called the Wiener algebra.) The sum of their squares will be as long as we work ... -1 A necessary condition for the convergence of a series \sum_n a_n is that the sequence (a_n) converges to 0 hence by the contraposition if (a_n) doesn't converge to 0 then the series diverges. For the given series we have$$\frac{(-1)^nn}{\sqrt{n^2+1}}\not\xrightarrow[n\to0]\;0$$so this series is divergent. 3 If a series \sum a_n converges, then a_n must go to zero. This is not the case in the example you posted, so the series cannot converge. By the way, the Leibniz test also requires the sequence to be monotonically decreasing. 0 Hint: Separate it into two subseries, for odd k's and for even k's. 0 If the sequence is uniformly convergent then we can find some n such that |f(t)-f_n(t)| < 1 for all t. Since f_n is bounded by, say, B, then we must have f be bounded by B+1. This has nothing to do with continuity, just uniform convergence and boundedness. 1 No. Consider the Taylor series for e^x around x=0. 0 Uniform convergence says: \forall \epsilon > 0, \forall n \geq N, such that |f_n(x) - f(x)| < \epsilon, \forall n \geq N and \forall x \in [0,1]. Then:$$|f_n(x)|=|f_n(x) - f(x)+f(x)| \leq \epsilon + f(x), \forall n \geq N \forall x \in [0,1]$$Finally a continous function over a interval is bounded pick for f that bound B and you ... 1 Start with the study of$$ 1-\frac nx\sin\frac xn $$on (0,2). It is F(\frac xn) where F(x) = 1-\frac 1x\sin x. As \lim_0 F = 0, for each \epsilon>0 one can find r>0 such as$$ |x|<r\Rightarrow |F(x)|<\epsilon $$then taking N>\frac 2r,$$n\ge N \ \ \&\ \ 0<x<2\Rightarrow \frac xn<\frac 2N<r$$and$$ \left|1-\frac ... 0 Let$f$be any nonnegative continuous function with support contained in$[0,1]$, and $$f_n(x)=nf(nx).$$ To get a positive function, just add something like$e^{-x^2}$to each$f_n$. 0 Take$f_n(x) = n$if$0\le x \le \frac 1n$,$E = [0,1]$and 0 otherwise.$f = 0$and$\int_E f_n = 1 \neq 0 = \int_E f$. 0 It follows from the very definition of (pointwise) convergence that $$\{\lim_{n \to \infty} X_n = X\} = \bigcap_{n \in \mathbb{N}} \bigcup_{m \in \mathbb{N}} \left\{ \sup_{k \geq m} |X_k-X| < \frac{1}{n} \right\}.$$ Hence, by the continuity of the measure$\mathbb{P}, \begin{align*} \mathbb{P}(\lim_{n \to \infty} X_n = X) &= \lim_{n \to \infty} ... 0 Hint: You need to make the proof in two steps. First step, the open halflines (sets of the form (-\infty, x) generate the Borel \sigma-algebra in \mathbb{R}. Second step, use the portmanteau theorem - which provides conditions that are equivalent to the convergence in distribution - as stated in Thomas's answer to prove the convergence in distribution. ... 2 A sequence X_1,X_2,\ldots of random variables is said to converge in distribution, or converge weakly, or converge in law to a random variable X if \lim_{n\to\infty}F_n(x)=F(x) $$for every number x\in\mathbb R at which F is continuous, where F_n(x)=\mathbb P(X_n\le x) and F(x)=\mathbb P(X\le x). Thus, we need to show that F(x) is ... 0 Start with the definition of what is the convergence. There is also a theorem (http://fr.wikipedia.org/wiki/Convergence_en_loi, can't find the english version) that says that you need to prove that P(X_n \in A) -> P(X \in A) for any A an open 0 Well, for a topological space X a set O \subset X is open iff for all x \in O, for any filter \mathcal{F} on X with \mathcal{F} \rightarrow x we have O \in \mathcal{F}. (Because O is a neighbourhood of X and a filter convergent to x in a topological space means that all neighbourhoods of x are in the filter.) So this comes down to: for ... 2$$ z^2-\sqrt{2}\,z+2=(z-a)(z-b)\text{ where }a=\frac{\sqrt2+\sqrt6\,i}{2},\ b=\frac{\sqrt2-\sqrt6\,i}{2}. $$Use partial fraction decomposition to write$$ \frac{1}{z^2-\sqrt{2}\,z+2}=\frac{A}{z-a}+\frac{B}{z-b}.$Now expand in power series$1/(z-a)$and$1/(z-b)$. 0 No. Take$z_n=-1$, so that$\prod_{n=1}^\infty |z_n|$trivially converges to$1$. But$z_n=1+(-2)$, and the definition of$\prod_{n=1}^\infty (1+(-2))$converging absolutely is that$\prod_{n=1}^\infty (1+|{-}2|) = \prod_{n=1}^\infty 3\$ converges, which it does not.

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