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0

Omran Kouba has correctly answered the OP question. Let's just see how the integral behaves as $n$ is great. As $n \rightarrow +\infty$, we have $$ \int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\:\mathrm dx = \frac{7\pi^4}{120}\frac{\ln n}{n^4}-\frac{45\:\zeta(5)}{2}\frac{\ln n}{n^5}+\mathcal{O}\left(\frac{\ln n}{n^6}\right) \tag1 $$ Recall that $$ ...


-3

You have a summary of relations between convergences on Wikipedia: http://en.wikipedia.org/wiki/Convergence_of_random_variables#Convergence_in_distribution "Properties" section There you can see that the a.s. convergence implies the convergence in distribution


4

We will prove that the limit is $+\infty$. Let $$\eqalign{I_n&=n^4\int_0^1\frac{x^n}{1+x^n}\ln^3 x\ln(1-x)dx\cr J_n&=n^4\int_0^1x^n\ln^3 x\ln(1-x)dx}$$ Clearly $$\frac{1}{2}J_n\leq I_n\leq J_n$$ because $\frac{1}{2}\leq\frac{1}{1+x^n}\leq1$ and $\ln^3x\ln(1-x)\geq0$ for $0<x<1$. So, let us consider $J_n$. We have ...


0

Another slightly different approach. We know that $ c_n - c_{n-1 } = \dfrac{0.01}{n} \implies c_n - c_{n- 1 } = \dfrac{0.01}{n} $ Let $ m \gt n \ge N $. Then, $$ |c_m - c_n| = |(c_m - c_{m- 1}) + (c_{m - 1} - c_{m- 2} ) + ... + (c_{n+ 1} - c_n)| $$ $$ = 0.01| \frac{1}{m} + \frac{1}{m -1} + ... + \frac{1}{n + 1} | $$ Especially, when $m = 2n$; $$ = ...


0

I think the easiest method is to use the Cauchy Criterion (for the first sequence). Given any $n \in \Bbb N$ use the suggestions below to create an $\epsilon \gt 0$ such that $|B(2n) - B(n)| \ge \epsilon$. Not that difficult to approximate. $$ |B(2n) - B(n)| = (n + 1/\sqrt{n^2+1})+.......2n/\sqrt{n^2+n} \ge \dfrac{n(n+1)}{\sqrt{n^2+n}} = \sqrt{n^2+n} \to ...


1

The comment box does not seem to like this equation! So if this is completely taking you in the wrong direction then please comment below and I will remove. Would this be of benefit here $$ \lim_{n\rightarrow \infty}\left[\lim_{\alpha\rightarrow 0,\beta\rightarrow 0}\dfrac{\partial^3}{\partial \alpha^3}\dfrac{\partial}{\partial ...


2

For the time being, the only thing I can tell you is that $$\begin{equation} I_n=n^4\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\,dx \end{equation}$$ seems to be an increasing function of $n$ (this has been done numerically): $$I_{10}=4.92397$$ $$I_{100}=18.2376$$ $$I_{1000}=31.7992$$ $$I_{10000}=44.97982$$ $$I_{100000}=58.0781$$


0

As Claude Leibovici said, the sequence does not converge. That is because the harmonic series $H_n=\sum\limits_{k=1}^n\frac{1}{k}$ diverges. We can see that because if we group the terms of the sum we see it is greater than an infinite sum of halves, which obviously diverges. The first term is 1, and we leave it stand-alone. The second one is a half, and we ...


0

Let us consider the more general case $$c_n = c_{n-1} + \frac{a}{n}, \ \ c_1 = b$$ the first terms of the sequence will be $b$, $b+\frac{a}{2}$, $b+\frac{5 a}{6}$, $b+\frac{13 a}{12}$ and more genrally $$c_n=b+a (H_n-1)$$ which generalizes what David Mitra commented. Then, just as its cousin, it does not converge.


0

No limit exists since this is not convergent. Because $1/n > \ln(1+1/n)$ (because $e^x>1+x$ when $x>0$.) Then $\sum_{n=1}^k 1/n > \sum_{n=1}^k \ln(n+1)-\ln(n) = \ln(k+1)\to\infty$


0

I only have taken a semester of undergraduate analysis, and what I know of measure theory has all been self taught. So take my solution as correct at your own risk. Let $\varepsilon>0$ be given. Since $m(E)<\infty$, let $m(E)=L$. Define $\delta=\frac{\varepsilon}{2+L}$. Since $f_n \rightarrow 0$ in measure, there is a positive integer $N$ so that ...


2

Convergence in $L^\infty$ does imply convergence in measure, even if $E$ has infinite measure. To see this, suppose that $\{f_n\}$ converges to $f$ in $L^\infty$. Let $\epsilon > 0$. Then there exists an $N_\epsilon$ such that $\|f_n - f\|_\infty < \epsilon$ for all $n > N_\epsilon$. Therefore, for all $n > N_\epsilon$, we see that $|f_n - f| ...


1

If you consider the largest integer $N$ in a tuple, there are a constant ($n$) number of places where it can occur, and all other entries can be $\leq N$. Over all tuples that have largest integer $N$ appearing, both the smallest and largest possible value of the terms in your series are on the order of $O(N^{-2\alpha})$ since $n$ is a constant. The total ...


1

Edit: Manoli, note that $n^{1/n} \lt 2$ for all $n$. It follows that $\dfrac{1}{n^{1+1/n}}\gt \dfrac{1}{2n}$. $$\sum_{n=1}^{\infty} \frac{1}{2n} \text{diverges}$$ so $\sum_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{n}}}$ also diverges.


2

First of all, $|z|<1$ is necessary, but maybe not sufficient (harmonic series). But here it is: \begin{align} \sum_{n=1}^\infty\left|\dfrac{1}{\frac{1}{z^n}-1}\right|=\\ \sum_{n=1}^\infty\left|\dfrac{z^n}{1-z^n}\right|\leq\\ \tag{1}\sum_{n=1}^\infty\dfrac{|z|^n}{1-|z|}=\\ \frac{1}{1-|z|}\sum_{n=1}^\infty|z|^n=\\ \left(\frac{1}{1-|z|}\right)^2<\infty\\ ...


1

See the definition in http://papers.nips.cc/paper/3646-on-the-convergence-of-the-concave-convex-procedure.pdf. The points generated by an algorithm with this property converge for any initial point. converge to a stationary point.


1

Suggestions: Recall that every weakly convergent sequence is bounded in the norm. Think of what it means for $n_{k}u_{k}$ to be bounded in the norm for every sequence $n_k$ Conclude that your sequence $u_k$ is very special.


-1

As far as I know optimisation would be the search for maxima/minima. If you'd imagine your function being a mountain chain, the optimum would equal a peak or a valley. To find the minima or maxima you could calculate the gradient of the function and starting from an initial point change the values of this initial point into the direction the gradient is ...


4

HINT: $$\frac{n!+1}{(n+1)!}=\frac1{n+1}+\frac1{(n+1)!}$$ The comparing series is $\sum\dfrac1n$


0

Global convergence is generally used in the context of Iterative numerical algorithms. It is usually defined as a sequence generated by the iterative algorithm converges to a solution point. This is important as it answers the question whether a particular algorithm, when initiated at a point far from the solution point, converges to it.


3

$\dfrac{\sqrt{n+2}-\sqrt{n-2}}{\sqrt n} = \dfrac{4}{\sqrt n(\sqrt{n+2}+\sqrt{n-2})}$ Compare it with $\sum \dfrac{1}{n}$


2

Hint: try multiplying the numerator and denominator by $\sqrt{n+2}+\sqrt{n-2}$.


0

Let $ A_n=\frac1n\,\begin {bmatrix}I_n\\0_n\end {bmatrix} $, $ B_n=\frac1n\,\begin {bmatrix}I_n\\I_n\end {bmatrix} $. Then $\|A_n-B_n\|=1/n $, while $$ P_{A_n}=\begin{bmatrix} I_n&0_n\\0_n&0_n\end {bmatrix}, \ \ \ P_{ B_n}=\frac12\,\begin {bmatrix} I_n& I_n\\ I_n& I_n\end {bmatrix}. $$ So $\|P_{A_n}-P_{B_n}\|=1/2$ for all $ n $.


3

By "the series $\sum\limits_n X_n$ converge almost surely", it is meant that there exists $\Omega'$ of probability $1$ for which if $\omega\in\Omega'$, then the series of real numbers $\sum\limits_{n=1}^{+\infty}X_n(\omega)$ is convergent. Define the event $E_n:=\{X_n\neq a_n\}$. We get by the assumption that $\mathbb P\left(\limsup\limits_nE_n\right)=0$. ...


1

Essentially, we have to investigate the behaviour of the series $$S=\sum_n (\log n\cdot r^n)^n,$$ where $r$ is a real number. Notice that if $|r|\lt 1$, then $\log n\cdot r^n$ is smaller than $1/2$ for $n$ large enough, hence the series $S$ is convergent. If $r=1$, the series diverges.


1

One knows that $nw^n\to0$ if and only if $|w|\lt1$ hence $n!\,w^{n!}\to0$ if and only if $|w|\lt1$ (a small argument is necessary for the "and only if" direction but you should be able to find it) thus the disk of convergence is centered at $___$ with radius $____$. Recall that the disk of convergence $D$ of a series $\sum\limits_na_nw^n$ is characterized ...


2

Almost. You need (a lot of) zero coefficients. Calling the function $f$, we have $$f(z) = \sum_{k=0}^\infty a_k\cdot z^k,$$ where $$a_k = \begin{cases} k &, k = n!\\ 0 &, \bigl(\forall n\bigr)(k \neq n!).\end{cases}$$ Now the Cauchy-Hadamard formula gives you the radius of convergence easily.


3

I do not know how much this could help you but $$M_n = \sum\limits_{i=1}^n\sum\limits_{j=1}^n\sqrt{ i^2 + j^2 } \lt \sum\limits_{i=1}^n\sum\limits_{j=1}^n(i+j)=n^2+n^3$$ So, as J.J. pointed it it out, the $n^3$ contribution seems to be clear. By the way, the limiting value, as answered by J.J., is $$\frac{1}{3} \left(\sqrt{2}+\sinh ^{-1}(1)\right)\simeq ...


1

We have $$ \sum_{j=1}^{n}\sqrt{i^2+j^2}\gt\sum_{j=1}^{n}j=\frac{n(n+1)}{2};i=1,2,\ldots,n $$ so $$ \sum\limits_{i=1}^n\sum\limits_{j=1}^n\sqrt{ i^2 + j^2 }\gt n\frac{n(n+1)}{2}=\frac{1}{2}(n^3+n^2) $$


3

Consider the following: $$M = \sum_{i=1}^n \sum_{j=1}^n \sqrt{i^2 + j^2} = n^3 \sum_{i=1}^n \sum_{j=1}^n \sqrt{\left(\frac{i}{n}\right)^2 + \left(\frac{j}{n}\right)^2} \cdot \frac{1}{n} \cdot \frac{1}{n}.$$ Now as $n \to \infty$ we have $$\sum_{i=1}^n \sum_{j=1}^n \sqrt{\left(\frac{i}{n}\right)^2 + \left(\frac{j}{n}\right)^2} \cdot \frac{1}{n} \cdot ...


0

Suppose not: then for some $\delta$ and some $n_k\uparrow\infty$, $$\lVert f_{n_k}-f\rVert_p\geqslant\delta,\quad k\geqslant 1.$$ Using the definition of convergence in measure, we can construct $m_k\uparrow\infty$ such that $\lambda\{x, |f_{n_{m_k}}(x)-f(x)|>2^{-k}\}\leqslant 2^{-k}.$ The sequence $(f_{n_{m_k}})_{k\geqslant 1}$ converges almost ...


0

We have $$\frac{1}{{\sin x}}:\frac{1}{x} = \frac{x}{{\sin x}} = \frac{1}{{\sin x/x}} \to 1$$ as $x\to 0^+$. In other word, $\frac{1}{\sin x}\sim \frac{1}{x}$ as $x\to 0^+$. But, it is well known that $\int_0^{1}\frac{1}{x}dx$ is divergent. So, we conclude that $\int_0^1 {\frac{1}{{\sin x}}dx}$ is divergent.


2

Set $x=y^2$, then $dx=2ydy$. So the integral becomes: $$I=\int_0^1 \frac {1}{(x+x^{5})^{1/2}}dx=\int_0^1 \frac {2y}{(y^2+y^{10})^{1/2}}dy=\int_0^1 \frac {2}{(1+y^5)^{1/2}}dy$$ Thus the integral is convergent.


1

$$\frac{1}{\sqrt{x+x^5}}=\frac{1}{\sqrt{x}\sqrt{1+x^4}}$$ Near $x\approx0$ you have $1+x^4 \approx 1$. And the integral of $1/\sqrt{x}$ converges near 0. Therefore your integral converges. Moreover, the result, as computed by Mathematica, is: $$\int_0^1\frac{\mathrm{d}x}{\sqrt{x+x^5}} = 2 \times\, _2 F_1\left(\frac{1}{8},\frac{1}{2};\frac{9}{8};-1\right) ...


2

Convergent. Compare with $\int_0^1 \dfrac{1}{x^{1/2}}\; dx$.


3

Outline: Note that $\sqrt{x+x^4}\ge x^{1/2}$ in our interval. Now recall that $\int_0^1 \frac{dx}{x^{1/2}}$ converges.


1

You're assuming the series $\sum\limits_{n=1}^\infty |a_n|$ converges. This, by definition means that the sequence of partial sums, $(S_m)$, given by $S_m=\sum\limits_{n=1}^m |a_n|$, converges. If you look closely, you should be able to see that your condition is just saying $(S_m)$ is a Cauchy sequence (which it is, of course).


4

You could use the following facts to find proper values of $A$ as well. Maybe it helps you for other integrands: Let $\lim_{x\to a^+}~(x-a)^pf(x)=A$. Then: If $p<1$ and $A$ is finite then $\int_a^bfdx$ converges. If $p\ge1$ and $A\neq0$ or $A=\infty$ then $\int_a^bfdx$ diverges. Now, you can see that why @Lucian's short hint can guide ...


3

Is this Convergent or Divergent ? Yes! :-) I have no Idea WHAT kind of function to compare with:/ Hint: $\sin x\sim x$ for $x\to0$.


0

HINT: The McLaurin expansión of $$F(x)=\int_0^x \frac{t^4}{(\sqrt{4+t^2})^3}\,dt$$ is given by: $$\begin{array}{rcl} F(x)&=&\sum_{k=0}^{\infty} \frac{F^{(k)}(0)}{k!}x^k\\ &=&F(0)+F'(0)x+\frac{F''(0)}{2}x^2+\ldots\\ &=&F'(0)x+\frac{F''(0)}{2}x^2+\ldots\\ \end{array}$$ Note that $F(0)=0$ because of the integration interval, and ...


1

Hint: Consider the function $f:\mathbb{R}^2\to\mathbb{R}$ defined by $$f(x,y) = \cases {0,&if $(x,y)=(0,0)$\\ xy^2/(x^2+y^4) &otherwise}$$ Is $f$ continuous when restricted to an arbitrary line in $\mathbb{R}^2$? Can you prove it? Is $f$ continuous? (Try taking a path along the graph of a non-linear polynomial. What is the simplest ...


3

There is probably a sign error in the problem, hence we consider $$Y_n=\left(1-\frac1n\right)^{n\bar X_n}-\mathrm e^{-\bar X_n}.$$ Note that, when $n\to\infty$, $$\log\left(1-\frac1n\right)=-\frac1n-\frac1{2n^2}+o\left(\frac1{n^2}\right),$$ hence $$\left(1-\frac1n\right)^{n\bar X_n}=\exp\left(-\bar X_n-\frac{\bar X_n}{2n}+o\left(\frac{\bar ...


0

Here is an answer for where to start: If you are unfamiliar with power series inversion, have a textbook or other table of these things handy. You should notice that the series for $\textrm{log}(1+x)$ near zero has two similarities with your series. 1) It is alternating 2) And has terms of the form $\frac{x^k}{k}$. I'll let you take it from there. You ...


0

try rewriting as $$ S = x^2 \sum_{n=0}^{+\infty} (\frac{-x^3}{2})^n $$


1

Promoting comment to answer: Both parts (a) and (b) are about $\|(x_n)\|$. In the first, you should calculate this norm and find when it's bounded. In the second, when it goes to zero. Keep in mind that these are sequences, not series. The sequence $n^p$ is bounded iff $p\le 0$, and goes to zero iff $p<0$. Your "$<-1$" comes from ...


1

This is the kernel of a continuous linear functional, hence closed. We have $|I(f_n)-I(f)|\leq\|f_n-f\|$, if $I(f_n)$ are all $0$ so is $I(f)$.


2

First observation: by the uniform boundedness principle, the sequence $(\lVert T^n\rVert)_{n\geqslant 1}$ is bounded. Let $x\in H$. Using compactness of $T$ and the fact that $T^nx\to 0$ weakly, we can find $n_k\uparrow \infty$ such that $\lVert T^{n_k}x\rVert\to 0$. Since $$\sup_{n_k\leqslant n\lt n_{k+1}}\lVert T^nx\rVert\leqslant \sup_{n_k\leqslant ...


0

I don't see why the mentioned equality yields the existence of the limits, but here is an alternative way to prove the (original) statement. It actually even shows $X_n \to X$ in $L^1$. Since $$\mathbb{E}|X_n| \to \mathbb{E}|X| \tag{1}$$ and $X \in L^1$, we have $X_n \in L^1$ for sufficiently large $n$ and $$|X_n-X| \leq |X_n|+ |X| \in L^1.$$ This ...


0

$\mathbb{E}|X_n| \to \mathbb{E}|X|$: No, this does not imply the statement. Simply consider $X_n := 1$ for $n$ odd and $X_n := -1$ for $n$ even. $\mathbb{E}X_n \to \mathbb{E}X$: Again, no. Consider $$X_n \sim \frac{1}{2} \delta_n + \frac{1}{2} \delta_{-n}.$$ With the additional condition: Consider the sequence ...


2

A standard counterexample is $f(x) = \begin{cases} e^{-{1 \over x^2}}, & x \neq 0 \\ 0, & x=0 \end{cases}$. The Taylor coefficients around $x=0$ are all zero, but the function is clearly not zero. (Note that this has nothing to do with the radius of convergence of the terms ${ f^{(k)}(a) \over k!}$, as in this case, all of these terms are zero.)



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