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0

You don't need Zorn's lemma. The function $f$ can be defined explicitly as $$ f_1\chi_{\{|x|\le 1\}}+\sum_{n=2}^\infty f_{n} \chi_{\{n-1<|x|\le n\}} $$ which is evidently in $L^1_{\rm loc}$ (on every bounded interval, only finitely many terms are nonzero). The convergence of $f_n$ to $f$ on every bounded interval follows from the fact that the restriction ...


0

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$ $\dfrac{df}{ds}=0$ , letting $f(0)=f_0$ , we have $f=f_0$ $\dfrac{dx}{ds}=v(f)=v(f_0)$ , letting $x(0)=F(f_0)$ , we have $x=F(f_0)+v(f_0)s=F(f)+v(f)t$


0

For $|x| > 1$, the sequence $x^{n!}$ is not even bounded, so the series does not converge. For $x = 1$, the sequence of partial sums is not bounded ($S_n = n$), so it does not converge. For $x = -1$, the sequence has periodic behavior ($S_{2n} = 1, S_{2n+1}=0$) and does not converge either. For $|x| < 1$ the series $\sum x^n$ converges absolutely ...


3

For the uniform convergence we have $$\sum_{n\geq1}e^{-nx+\cos\left(nx\right)}\leq\sum_{n\geq1}e^{-nx}\leq\sum_{n\geq1}e^{-na}=\frac{1}{e^{a}-1} $$ then we have uniform convergence by M-test. For the differentiation, we have to prove that $$f'\left(x\right)=\sum_{n\geq1}f_{n}'\left(x\right) $$ and so the uniform convergence of ...


1

Your proof is fine, but I added some points and shortened it. You can say that $f_n \rightarrow f$ pointwise, where $f = \begin{cases} 0 & \text{if } x = 0 \\ \frac{x}{\sin(x)} & \text{if } x \in (0, \frac{\pi}{2}] \end{cases}$. For $x = 0$, $\lim_{n \rightarrow \infty}f_n(0) = 0$ and For $x \in (0,\frac{\pi}{2}]$, $\lim_{n \rightarrow ...


1

Unfortunately there is one problem in your reasoning. Note that $\int f_{n}>0 $ is perfectly fine and one might still have $f_{n}\to 0$ almost surely and $\int f_{n}\to 0$. Take for example $f_{n}=\chi_{[0,\frac{1}{n}]}$ for all $n$, i.e. $f_{n}=1$ on $[0,\frac{1}{n}]$ and zero otherwise. Then $\int f_{n} = \frac{1}{n}>0$ for all $n$ but $f_{n}\to 0$ ...


1

Consider $(\mathbb{Z},\mathcal{P}(\mathbb{Z}))$ endowed with the counting measure $$\mu(B) := \sum_{h \in \mathbb{Z}} \delta_h(B).$$ Then for any integrable function $f \in L^1$, we have $$\int f \, d\mu = \sum_{h \in \mathbb{Z}} f(h). \tag{1}$$ Define $$f_n(h) := \begin{cases} \left(1- \frac{|h|}{n} \right) \gamma(h), & |h| < n, \\ 0, & |h| ...


0

One good option here is the logarithmic test: compute $\lim \limits _{n \to \infty} \frac {\log \frac 1 {x_n}} {\log n}$. If you get a number $<1$, the series is divergent; if you get $>1$, it is convergent; if you get $1$, you cannot draw any conclusion and you must try some other method. The test works only for series with positive terms. In your ...


2

I'm assuming you aim to determine the convergence of $$\sum_{n=2}^\infty \frac{1}{\sqrt{n}\log(n)}$$ Let's try to find a series that is smaller than this one, but still divergent. For example, $$\sum_{n=2}^\infty \frac{1}{n\log(n)}$$ By the integral test we have $$\int_{2}^\infty \frac{1}{x\log(x)}\text{d}x = \log(\log(x))\Big\vert^\infty_2 = \infty $$ so ...


0

The Riemann-Lebesque Lemma says that if $g\colon[a,b]\to\mathbb{R}$ is Riemann integrable, then $$ \lim_{\lambda\to\infty}\int_a^bg(x)\cos(\lambda\,x)\,dx=\lim_{\lambda\to\infty}\int_a^bg(x)\sin(\lambda\,x)\,dx=0. $$ Under the conditions on $f$, the function $h$ is Riemann integrable on $[-\pi,\pi]$. This implies that $$ \lim_{n\to\infty}\int_{-\pi}^\pi ...


0

For b) with ratio test: $$ \lim\limits_{n \rightarrow \infty}{\frac{cosh(n)}{cosh(n+1)}} = \lim\limits_{n \rightarrow \infty}{\frac{e^k+e^{-k}}{e^{k+1}+e^{-k-1}}} = \lim\limits_{n \rightarrow \infty}{\frac{1+e^{-2k}}{e+e^{-2k-1}}} =\frac{1}{e} $$


0

To do b) with the ratio test, consider that the ratio between terms is $$\frac{\cosh n}{\cosh(n+1)} = \frac{e^n+e^{-n}}{e^{n+1}+e^{-n-1}}\to e^{-1}<1$$ The hard part is rigorously showing the convergence but the idea is that the $e^{-n}$ terms become negligible as $n\to\infty$. EDIT: This isn't hard. Because $\lim_{n\to\infty}e^{-n}=0,$ ...


2

For (a) you do not need the ratio test: just note that $\lim \limits _{n \to \infty} \frac {\sinh n} {\mathbb e ^n} = \lim \limits _{n \to \infty} \frac {\mathbb e ^n - \mathbb e ^{-n}} {2 \mathbb e ^n} = \frac 1 2$. By the zero test, since the limit is not $0$ then the series diverges. For (b) indeed, one could use the ratio test, but I shall use the root ...


1

What if you define $x=e^{-a}$ ? This would make $$S(a)=\sum_{n=1}^\infty n e^{-n a}=\sum_{n=1}^\infty n x^n=x \sum_{n=1}^\infty n x^{n-1}=x \frac{d}{dx}\Big( \sum_{n=1}^\infty x^n\Big)$$ You have then a classical summation; differentiate it to get $$S(a)=\frac{x}{(1-x)^2}=\frac{e^a}{\left(1-e^a\right)^2}$$ which undefined only for $a=0$; you could also note ...


2

Another easy test you can use: the ratio or d'Alembert test. $$\frac{a_{n+1}}{a_n}=\frac{(n+1)e^{-(n+1)a}}{ne^{-an}}=\frac{n+1}n\cdot\frac1{e^a}\xrightarrow[n\to\infty]{}\frac1{e^a}<1$$


1

we know that sigma n x^n converges to x/(1-x)^2 for abs(x)<1. This is easy to show by ratio or root test. our series is the same as sigma n(1/e^a)^n,0<1/e^a<1, since a>0. In fact we can fond the sum.


1

You can also use a comparison test by noticing that $ne^{-na}=O\left(\frac{1}{n^2}\right)$ for $a>0$.


2

I think that the best way is to use the root test:http://en.wikipedia.org/wiki/Convergence_tests. Indeed you have $(ne^{-an})^{\frac{1}{n}}\rightarrow e^{-a}<1$.


0

By ratio test you can find that the series converges if $\left|x\right|<1 $. In this case this series admits a closed form in terms of special functions. In fact we have $$\sum_{n\geq1}\frac{x^{n}}{\sqrt{n+1}}=x\sum_{n\geq0}\frac{x^{n}}{\sqrt{n+1}}=x\Phi\left(x,\frac{1}{2},x\right) $$ where $\Phi\left(x,s,a\right) $ is the Lerch Trascendent. For $x=1 ...


0

An often easier way is that the radius of covergence is the least upper bound of the $r\ge 0$ such that $a_nr^n\to 0$ as $n\to\infty$. Here, $\,\dfrac{a_n}{r^n}=\Bigl(\dfrac r2\Bigr)^n$ and it is well known it tends to $0$ if and only if $\,\dfrac r2<1\iff r<2$, hence $R=2$.


0

Split this up into three parts: 1). $x\in (-1,1)$ 2). $x\notin (-1,1)$ 3). $x=1$ or $x=-1$. 1). In this case your general term is $\leq $ $x^n$. On comparison with the geometric series, your series then converges. 2) and 3). In these cases, the series diverges, because the general term does not go to zero as $n\rightarrow \infty$.


2

To complement the answer by CPM: You can also use the formula of Chauchy-Hadamard. It states, that the series $\sum a_n (x-c)^n$ has the radius of convergence $$R=\frac{1}{\lim \sup_{n\to\infty} \sqrt[n]{|a_n|}}$$ Thus $$R=\frac{1}{\lim \sup_{n\to\infty} \sqrt[n]{\frac{1}{2^n}}}=\frac{1}{\lim \sup_{n\to\infty} \frac{1}{2}} = 2$$


2

The center of your interval of convergence is $x=1$, but not the radius. To find the radius of convergence you need to use the ratio test. $$\lim_{x\to \infty}\left| \frac{(x-1)^{n+1}}{2^{n+1}}\cdot \frac{2^{n}}{(x-1)^{n}}\right|=\lim_{x \to \infty} \left|\frac{x-1}{2}\right| $$ To converge, this needs to be less than $1$. Thus $|x-1|<2$. So your ...


2

From Calculus: 8th Edition by Larson: [A]n infinite series of the form $$ \sum_{n = 0}^\infty a_n(x-c)^n$$ is called a power series centered at c, where c is a constant. So here $c = -4$.


2

Your use of the ratio test is fine--clearly, it won't converge outside of $[-1,1]$, and you are correct in asserting that it doesn't converge for $x=-1$ and $x=1$. You probably still have to prove that it converges for $x\in (-1,1)$, but you can use the comparison test against the series $a_n = x^n$ for that.


4

$$ f(x) = \lim_{n\rightarrow \infty} \frac{nx}{1+n\sin(x)} = \lim_{n\rightarrow \infty}\frac{x/\sin(x)}{\frac{1}{n\sin(x)}+1} = \frac{x}{\sin(x)} \quad (x \neq 0) $$ and $$ f(0) =\lim_{n\rightarrow \infty} \frac{n(0)}{1+n\sin(0)} = 0 $$ but $$ \lim_{x\rightarrow 0+} f(x)=1$$ so $f$ is not continuous at $0$. The $f_n$ are all continuous at $0$ and if the ...


0

When $x=0$ then $f(x)=0$. When $x\not = 0$ then $f(x)=\frac{x}{\sin x}$. So, $$f(x)=\begin{cases}0 & \text{ for } x=0\\\frac{x}{\sin x} & \text{ for } x\in (0,\pi/2]\end{cases}$$As, $f$ is not continuous at $x=0$, so $\{f_n(x)\}$ is NOT uniform convergent.


1

The reason this test is inconclusive is that even two series with exactly the same successive ratios can have different convergence properties when the limit of the successive ratios are 1. For example, the Harmonic series $\sum 1/n$ diverges, but the alternating harmonic series, $\sum (-1)^n 1/n$ converges. For both of these, the ratio test yields ...


1

Assuming $\epsilon$ is a vector and that "$\ge$" must be considered component-wise. $\mathbb{P}(X^n\ge\epsilon)=\mathbb{P}\left(\bigcap_{i=1}^m\{X^n_i\ge \epsilon_i\}\right)\le\min_{i=1}^m\mathbb{P}(X^n_i\ge \epsilon_i)$, so the if part is true. It actually holds the stronger $$\exists j\ \liminf_n \mathbb{P}(X^n_j\ge\epsilon_j)=0\Longrightarrow ...


2

No, there isn't. As long as you can apply the integral test, then it has to work: if $f : [0, \infty) \to [0, \infty)$ is decreasing, then the series $\sum f(n)$ converges iff the improper integral $\int_0^\infty f(t) dt$ converges. This isn't, say, like the root test or the ratio test that can be inconclusive, even when you can apply it (when $\lim ...


0

Let $\epsilon > 0$ be given. Notice that $\{x_n\} \rightarrow x$ and $\{y_n\} \rightarrow y$ implies there exists $N_1,N_2 \in \mathbb{N}$ such that $|x_n - x| < \frac{\epsilon}{2}$ for all $n > N_1$ and $|y_n - y| < \frac{\epsilon}{2}$ for all $n > N_2$. Let $N = max\{N_1,N_2\}$ then we have that $|x_n + y_n - (x+y)| = |(x_n - x) + (y_n - y)| ...


1

hint: $|(x_n+y_n)-(x'+y')| \leq |x_n-x'| + |y_n-y'|$


0

$(-1)^{n!} = 1$ for $n \ge 2$ (since $n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1$ is even for $n \ge 2$), so $$\lim_{n \to \infty}(-1)^{n!} = 1,$$and$$\sum_{i=0}^\infty (-1)^{n!} = (-1) + (-1) + 1 + 1 + 1 + \dots$$does not converge.


3

For $n\geq 2$, note $n!$ is even so $(-1)^{n!}=1$. Thus this converges to $1$.


0

Assume such a $K$ exists. Let $\varepsilon>0$, we aim to find $N$ such that $\lVert s_m-s_n\rVert<\varepsilon$ whenever $m,n\geq N$. We have $\sum_{j=1}^n\lVert x_j\rVert^2\leq K$ for each $n$, so the series $S:=\sum_{j=1}^\infty \lVert x_j\rVert^2$ converges and has $S\leq K$ (as it is a bounded, increasing sequence of positive real numbers). Now ...


1

No, continuity on $\bigcup_{n \in \mathbb{N}} \text{Img}(X_n)$ is not enough. (Counter)Example: Consider $([0,1],\mathcal{B}[0,1])$ endowed with the Lebesgue measure, $$g(x) := 1_{\{0\}}(x)$$ and $$X_n := \frac{1}{n}.$$ Then $X_n \to X := 0$ almost surely, $g$ is continuous on $(0,\infty) \supseteq \bigcup_n \text{Img}(X_n)$, but $g(X_n)=0$ does not ...


0

As you say, by the ratio test the radius of convergence as a power series about $0$ is indeed $R=1$, so the series must converge in $D=\{z \in \mathbb{C} : |z|<1 \}$. Lets see what happens if $z \in \partial D=\{ z: |z|=1\}$. Consider the set $$ E=\{ e^{2\pi i \cdot r} : r \in \mathbb{Q} \} \subset \partial D $$ For every $z \in E$, we have $$ ...


1

You have already proved that $$\lim_{n\to\infty} f_n(x) = \begin{cases}0 & x=0\\ \frac x{\sin x} & x \ne 0 \end{cases}$$ And this limit is indeed pointwise (as you can fix $x\ne 0$ and obtain the limit $\frac x{\sin x}$ and fix $x=0$ and obtain $f_n(0) = 0 \to 0$). Furthermore because the $f_n$ are continuous for $x\in [0, \frac\pi2]$ and the limit ...


0

Hint: Leibniz series, therefore it is convergent. To show, that it is Leibniz: $|a_n| \rightarrow 0$, as $n \rightarrow \infty$, that is true, since $| \frac23|^n$ will approach to $0$ as $n \rightarrow \infty$. it must be alternating, that is easy to see, since if $n \equiv 0$ (modulo $2$) $\rightarrow $ $a_n > 0$, and else, $a_n < 0$. |$a_n$| must ...


4

As you have guessed, the answer is NO in general. And a single counter example is enough: take $\{X_n\}$ i.i.d. uniform $(0,1)$ then $|X_n|< n$ for all $n\geq 1$ so that $$ Y_n=X_n\implies\text{Var}{Y_n}=\text{Var}{X_n}=\frac{1}{12}\implies\sum_{n=1}^\infty\frac{\text{Var}{Y_n}}{n}=\frac{1}{12}\sum_{n=1}^\infty\frac{1}{n}=\infty. $$


1

Note that \begin{equation} \begin{split} |u_\epsilon (x) - u(x)| &= \left|\int_B u(y) \eta_\epsilon (y-x) dy - u(x) \right|\\ &= \left|\int_B (u(y) - u(x)) \eta_\epsilon(y-x) dy \right|\\ \end{split} \end{equation} But the support of $\eta_\epsilon$ is so small, so the integration is over where $|y-x|$ is small. (Then try to use the uniform ...


1

Define $$A_n := \left\{ \left| \frac{X_n}{n} \right| \geq 1 \right\}.$$ The set $$A := \limsup_{n \to \infty} A_n$$ satisfies, by the Borel-Cantelli lemma, $\mathbb{P}(A)>0$ and $\frac{X_n(\omega)}{n}$ does not converge to $0$ for each $\omega \in A$.


1

This reasoning is incorrect. The issue is that there are series that converge with terms that are all non-zero. A classical example is the geometric series. Define $a_n = ( \frac{1}{2} )^n$ for $n \geq 0$. It is clear that there are no solutions to $a_n = 0$, and yet the series converges! (in this case, $\sum a_n = 2$ !) Moreover, what you have written ...


2

No this reasoning is not correct. The terms of the geometric sequence $a_n = \dfrac{1}{2^n}$ are never equal to zero but $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{2^n}$ converges to $1$. Also, all the terms of this geometric series are distinct. So, for any $x$ there is at most one solution $n$ to $a_n = x$. Thus, your second line of reasoning is also ...


1

For 1), recall that $\mathbb P=\delta_0$ just means that $\mathbb P(\{0\})=1$. So $X_n(0) = (1-0)^n = 1$, and $\mathbb P(X_n=1)=1$. Hence trivially $X_n\stackrel{a.s.}{\longrightarrow}1$ and $X_n\stackrel{d}{\longrightarrow}1$ For 2), recall that $\mathbb P=\frac12\delta_0 + \frac12\delta_1$ means $\mathbb P(\{0\})=\mathbb P(\{1\})=\frac12$. So $X_n(0) = 1$ ...


2

Hints: Fix $\epsilon>0$. Fix $k \in \mathbb{N}$. Using Markov's inequality, show that $$A_N^k := \left\{x; \exists n \geq N: |f_n(x)-f(x)| \geq \frac{1}{k} \right\}$$ satisfies $$m(A_N^k) \leq k \sum_{n=N}^{\infty} \|f_n-f\|_{L^1}.$$ Conclude from the first step that there exists $N=N(k)$ such that $m(A_{N(k)}^k) \leq \epsilon 2^{-k}$. Set $$A := ...


1

The general answer to your question is that it doesn't converge to zero for $\alpha \leq 1$. Consider the case $\alpha = 1$. This is the Law of Large Numbers. If your random variables were iid then that converges to the mean which may not be zero. For $\alpha > 1/2$ we have the sum being even larger. And in the case $\alpha = 1/2$ (which you didn't ...


1

The point is that we want to show that we can get $\|x_n\|$ as close to $\|x\|$ as we wish based on the assumption that we can get $\|x_n - x\|$ as close to $0$ as we wish. The bound $|\|x_n\| - \|x\|| \leq \|x_n -x\|$ guarantees this, because we can just take the limit on both sides: $$\lim_{n \to \infty} |\|x_n\| - \|x\|| \leq \lim_{n \to \infty} \|x_n - ...


3

Consider $A = \{ 2n : n \in \mathbb{N} \}$ and $B = \{ 2n-1 : n \in \mathbb{N} \}$. Note that $B = \mathbb{N} \setminus A$, and both $\sum_{a \in A} \frac{1}{a}$ and $\sum_{b \in B} \frac{1}{b}$ diverge. Therefore neither $A$ nor $B$ are convergent sets, and so by definition of the topology they are not closed. As a set is open iff its complement is closed, ...


1

Your first inequality is invalid, if you make the (positive) denominator smaller, since the numerator is also positive, you make the fraction larger. Actually, the series is normally convergent on $[a,+\infty)$ for all $a > 0$, since for $n \geqslant 2$ $$0 < U_n(x) = \frac{x}{(1+(n-1)x)(1+nx)} < \frac{x}{(n-1)xnx} = \frac{1}{n(n-1)x} \leqslant ...



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