# Tag Info

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 ...

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 $$Y_n=X_n\implies\text{Var}{Y_n}=\text{Var}{X_n}=\frac{1}{12}\implies\sum_{n=1}^\infty\frac{\text{Var}{Y_n}}{n}=\frac{1}{12}\sum_{n=1}^\infty\frac{1}{n}=\infty.$$

3

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

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 ...

2

Once you have split the fraction up, write down the first few terms of the initial sum in the split form. Can you find an explicit form for the sum to $N$? Then say what happens as $N\to \infty$? What you have done is to split a sum with positive terms into one with both positive and negative terms. This second sum can only be safely rearranged as you have ...

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

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 $$A := ... 2 I'm assuming you aim to determine the convergence of$$\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 $$so ... 2 For (a) you do not need the ratio test: just note that \lim \limits _{n \to \infty} \frac {\sinh n} {\mathbb e ^n} = \lim \limits _{n \to \infty} \frac {\mathbb e ^n - \mathbb e ^{-n}} {2 \mathbb e ^n} = \frac 1 2. By the zero test, since the limit is not 0 then the series diverges. For (b) indeed, one could use the ratio test, but I shall use the root ... 2 No this reasoning is not correct. The terms of the geometric sequence a_n = \dfrac{1}{2^n} are never equal to zero but \displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{2^n} converges to 1. Also, all the terms of this geometric series are distinct. So, for any x there is at most one solution n to a_n = x. Thus, your second line of reasoning is also ... 2 Unfortunately there is one problem in your reasoning. Note that \int f_{n}>0  is perfectly fine and one might still have f_{n}\to 0 almost surely and \int f_{n}\to 0. Take for example f_{n}=\chi_{[0,\frac{1}{n}]} for all n, i.e. f_{n}=1 on [0,\frac{1}{n}] and zero otherwise. Then \int f_{n} = \frac{1}{n}>0 for all n but f_{n}\to 0 ... 2 Fix \epsilon>0. Since X is integrable, there exists R>0 such that$$\int_{|X| \geq R} |X| \, d\mu < \epsilon.$$Thus,$$\begin{align*} \left| \int_{A_n} X \, d\mu - \int_A X \, d\mu \right| &= \left| \int (1_{A_n}-1_A) X \, d\mu \right| \\ &\leq 2 \int_{|X| \geq R} |X| \, d\mathbb{P} + \int_{|X|<R} |1_{A_n}-1_A| \cdot |X| \, d\mu ...

2

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 ... 2 Another easy test you can use: the ratio or d'Alembert test. $$\frac{a_{n+1}}{a_n}=\frac{(n+1)e^{-(n+1)a}}{ne^{-an}}=\frac{n+1}n\cdot\frac1{e^a}\xrightarrow[n\to\infty]{}\frac1{e^a}<1$$ 2 I think that the best way is to use the root test:http://en.wikipedia.org/wiki/Convergence_tests. Indeed you have$(ne^{-an})^{\frac{1}{n}}\rightarrow e^{-a}<1$. 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 ...

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.

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 $X_n(0) = 1$ ...

1

Note that $$\begin{split} |u_\epsilon (x) - u(x)| &= \left|\int_B u(y) \eta_\epsilon (y-x) dy - u(x) \right|\\ &= \left|\int_B (u(y) - u(x)) \eta_\epsilon(y-x) dy \right|\\ \end{split}$$ 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

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 ...

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 $a>0$.

1

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

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 ...

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

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

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 ...

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 - ...

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