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0

Question One. Let \begin{align} J=\int_0^1y \left( \left( 1+\frac{1}{y^2} \right)\log (1+y^2) -1 \right)dy&= \int_0^1 \frac{1}{y}\left( y^2+1 \right)\log (1+y^2)dy - \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,( *) \end{align} Setting \begin{align} I= \int_0^1 \frac{1}{y}\left( y^2+1 \right)\log (1+y^2)dy \end{align} ...

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Just an addendum to Marco's answer (the answer for question 2): $$J(\alpha,\beta)=\iint_{(0,1)^2} x^{\alpha}y^{\beta}\text{arctanh}(xy)\,dx\,dy = \sum_{n\geq 0}\frac{1}{(2n+1)(2n+2+\alpha)(2n+2+\beta)}$$ can be computed through partial fraction decomposition and: $$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}.$$ Namely, $$... 2 The integral of the question 1 is not difficult to evaluate. We have$$\int_{0}^{1}\left(\left(y+\frac{1}{y}\right)\log\left(1+y^{2}\right)-y\right)dy=\int_{0}^{1}y\log\left(1+y^{2}\right)dy+\int_{0}^{1}\frac{\log\left(1+y^{2}\right)}{y}dy-\int_{0}^{1}ydy. $$For the first integral take y^{2}+1=u  to get ... 1 Your approaches are both correct, you just have to finish them. First try: this is the easiest method:$$\lim \limits _{n \to \infty} \frac{\varphi \left( \frac 1 n \right) - \varphi \left( - \frac 1 n \right)} {\frac 1 n} = \lim \limits _{n \to \infty} \frac{\varphi \left( \frac 1 n \right) - \varphi (0)} {\frac 1 n} + \lim \limits _{n \to \infty} ...

1

Dirichlet's test is enough: $\{x\}-\frac{1}{2}$ is a function with a bounded primitive and $\frac{1-s\log x}{x^{s+1}}$ is monotonically convergent to zero from some point on.

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The distribution $f_a$ was defined for every $a>0$ in Strichartz's book by $$\langle f_a , \varphi \rangle = \int_{-\infty}^{-a} \frac{\varphi(x)}{|x|}\, dx + \int_a^{+\infty} \frac{\varphi(x)}{|x|}\, dx + \int_{-a}^a \frac{\varphi(x)-\varphi(0)}{|x|}\, dx.$$ If $\varphi(0)=0$, then $$\langle f_a , \varphi \rangle = \int_{-\infty}^{-a} ... 2 For \;|x|<1\; :$$\frac1{1-x^2}=\sum_{n=0}^\infty x^{2n}\stackrel{\text{diff.}}\implies\frac{2x}{(1-x^2)^2}=\sum_{n=1}^\infty 2n\,x^{2n-1}$$Now just do a little cosmetics to the above and get your answer. 3 The series converges: Using \sin x = x - \frac{x^3}{6} + o(x^4) around 0, we get that when n\to \infty$$ \sin\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n\ln^2(n)}\right) = \underbrace{\frac{\sin(3n)}{\sqrt{n}}}_{a_n} + \underbrace{\frac{1}{n\ln^2(n)} + O\left(\frac{1}{n^{3/2}}\right)}_{b_n}. $$The series \sum_{n=2}^\infty (-1)^n b_n converges ... 0 Let L_1, L_2 be different limits of subsequences. Let \epsilon = {1 \over 3} |L_1-L_2|. Then B(L_1,\epsilon), B(L_2,\epsilon)each contain an infinite number of points of the sequence. In particular, the sequence cannot be Cauchy. 1 First, note that since \sin x - x \le Cx^3 for |x|\le1 and some constant C, we have \begin{multline*} \bigg| \sum_{n=2}^\infty \bigg\{ (-1)^n \sin \bigg( \frac{\sin3n}{\sqrt n} + \frac1{n\log^2n} \bigg) - (-1)^n \bigg( \frac{\sin3n}{\sqrt n} + \frac1{n\log^2n} \bigg) \bigg\} \bigg| \\ \le C \sum_{n=2}^\infty \bigg| \frac{\sin3n}{\sqrt n} + ... 1 Write \sin x=x-x^3/6+x^3\varepsilon(x), where \varepsilon is such that \lim_{x\to 0}\varepsilon(x)=0. Consequently,$$ (-1)^n \sin\left( \frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n (\ln n)^2}\right)=(-1)^n\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n (\ln n)^2}\right)-\frac{(-1)^n}6\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n (\ln n)^2}\right)^3\\ ...

3

Hint: By mean value theorem, for each $n\in\mathbb{N}$ there exists $\xi_n\in(2n,2n+1)$ such that $$\log(2n+1)-\log(2n)=\frac{1}{\xi_n}\geq\frac{1}{2n+1}\geq\frac{1}{3n}$$

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Yes, there is. As it is a series with positive terms, you can use equivalents: $$\log(2n+1)-\log(2n)=\log\Bigl(1+\frac1{2n}\Bigr)\sim_\infty\frac1{2n},$$ which diverges.

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HINT: $$\sin(n)\sin\left(\frac{(-1)^n}{n^{1/4}}\right)=(-1)^n\sin(n)\sin\left(\frac{1}{n^{1/4}}\right)$$ Now, show that there exists a number $M$ (added hint: $M=\sec(1/2)$ suffices) such that for all $N$ $$\left|\sum_{n=1}^N (-1)^n\sin(n)\right|\le M$$ Finally, apply Dirichlet's Test noting that $\sin\left(\frac{1}{n^{1/4}}\right)>0$ and ...

2

Let $f_n(x) = e^{-x^2} n \sin\left(\dfrac{x}{n}\right)\cdot \chi_{[0,n^2]}(x)$. On the interval $[0,1]$, this is a monotone increasing sequence of functions, and MCT can be applied. On the interval $[1,\infty)$, you have $$\int^{n^2}_1 e^{-x^2} n \sin\left(\frac{x}{n}\right) \, dx = \int\limits_{[1,\infty)} f_n(x)\,dx$$ and $$|f_n(x)| \le xe^{-x^2} ... 0 The terms have to be (well) defined, that's not the case with your 'first term', f(1). You can always split the terms of a sequence into a finite first part (say the first k terms) and the "tail" (all the others). If a sequence converges, it has a limit L \in \mathbb{R}. Because it converges, we know there is a number N such that all the terms ... 0 Your "sequence" does not diverge at n=1, it is simply not defined at that point, so your domain would be \mathbb{N} \backslash \{1\}. If you say that something diverges you mean that it "tends towards" infinity and not that it actually "is infinity" (whatever that's supposed to mean). The theorem is easily seen, as you can not "reach infinity" in finitely ... 1 Look separately at the intervals N^2\le n < (N+1)^2, with N\gg 1. For these n, we have that$$ |\sin (n+\log n) - \sin (n+\log N^2)| \le \log (n/N^2) \lesssim 1/N \lesssim 1/n^{1/2} , $$and since \sum n^{-3/2}<\infty, we may replace \sin (n+\log n) by \sin (n+\alpha_N), for the n currently under consideration. However, as you already ... 1 Another fun way of doing it is by means of a generating function. Define$$ A(t) = \sum_{n\geq 0} x_nt^n = x_0 + x_1 t + x_2 t^2 + \dots \quad (1) $$for real t<1. Then multiply your recursion formula by t and sum to get$$ \sum_{n\geq 0} x_{n+1} t^n = \frac{1}{2} \sum_{n\geq 0} x_n t^n + \frac{1}{2} \sum_{n\geq 0} t^n. $$Note that the left-hand ... 3 Without too many details: |(1+i/n)^n| = (1 + \frac{1}{n^2})^{n/2} \arg(1 + i/n)^n = n\tan^{-1} 1/n as n \to \infty, the modulus goes to 1 and the argument goes to 1. This is exactly e^i. 0 If e.g. f is prescribed by x\mapsto0 and f_n by x\mapsto\frac1{n}\arctan x then:$$\sup_{x\in\mathbb R}|f_n(x)-f(x)|=\frac{\pi}{2n}$$while for every x\in\mathbb R:$$|f_n(x)-f(x)|<\frac{\pi}{2n}$$So there is no specific x_0\in\mathbb R with |f_n(x_0)-f(x_0)|=\sup_{x\in\mathbb R}|f_n(x)-f(x)| here. In order to prove for some ... 1 No. Let B=c_0 and let (e_n) be the standard basis. Let \alpha_j=1. Then ||p_n||=1 but p_n does not converge. 0 Here's one possible process for discovering the general formula: try applying the recurrence a few times:$$ \begin{aligned} x_1&=\frac{x_0+1}{2}=\frac{x_0}{2}+\frac{1}{2}\\ x_2&=\frac{x_1+1}{2}=\frac{x_1}{2}+\frac{1}{2}=\frac{x_0}{4}+\frac{1}{4}+\frac{1}{2}=\frac{x_0}{4}+\frac{3}{4}\\ ...

0

The integral is convergent by Dirichlet's test, since $\cos x$ has a bounded primitive while $\frac{1}{\sqrt[3]{\ln x}}$ is decreasing towards zero. Integration by parts gives: $$\int_{e}^{M}\frac{\cos x}{\sqrt[3]{\log x}}\,dx = \left.\frac{\sin x}{\sqrt[3]{\ln x}}\right|_{e}^{M}+\int_{e}^{M}\frac{\sin x}{3x\ln x\sqrt[3]{\log x}}\,dx$$and $$... 0 Hint: It's not too difficult to show (by induction if you want) that if x_{n+1}=ax_n +b for all n \geq 0 and a\not=1, then$$x_n = a^n\left(x_0-\frac{b}{1-a}\right)+\frac{b}{1-a}.$$Then, in your case a=\frac{1}{2}=b. Therefore x_n= (\frac{1}{2})^n(x_0-1)+1. 1 Martin gave a beautiful answer to your question but let me take this opportunity to advertise a solution to a more general question of which C*-algebras may be written as tensor products of two infinite-dimensional C*-algebras. In particular, it is proved that B(\ell_2) as well as the Calkin algebra cannot be decomposed like that no matter which tensor ... 0 Note that \sin^4(t) + \sin^4(t+1) is continuous and nonnegative, is never 0 and is periodic. It follows that there is a constant c> 0 such that \sin^4(t) + \sin^4(t+1) > c for all t. Also note that \sin^4(n)/\ln (n+1/n) \ge \sin^4(n)/\ln (2n). So it's enough to show the series \sum \sin^4(n)/\ln (2n) = \infty. Here observe that$$ ...

3

Note that $$\sin^4 n = (\sin^2 n)^2 = \left( \frac{1 - \cos 2n }2 \right)^2 = \frac{1 - 2\cos 2n + \cos^2 2n}4$$ $$= \frac{1-2\cos 2n}4 + \frac{1+2\cos 4n}8 = \frac38-\frac{\cos 2n}2+\frac{\cos 4n}4.$$ We split the sum into three parts: $$\sum_{n=1}^{\infty} \frac{\sin^4 n }{\ln(n+\frac1n)} = \sum_{n=1}^{\infty} \frac38 \frac1{\ln(n+\frac1n)} - ... 3 First, note that there exists \delta such that |\sin^4(x)|<\delta implies |\sin^4(x+1)|>\delta. To see this, draw a circle. The angles x for which |\sin(x)|<\delta correspond so arcs around (1,0) (at angle 0) and (-1,0) (at angle \pi), so if these arcs are small enough, which corresponds to \delta being small, then adding 1 to ... 3 Fine as they are, the other answers gloss over an important nuance: In a comparison of (positive) sequences, if a_n\le b_n and \lim_{n\to\infty}b_n=L, we cannot conclude that \lim_{n\to\infty}a_n=L. In fact, we cannot even conclude that \lim_{n\to\infty}a_n exists (unless L=0). We need to squeeze a_n between two sequences that both converge to ... -1 Factor n^2 out of the numerator and denominator to get$$a_n = \sum_{k=1}^n\frac{2 + \frac{2^k}{n} + \frac{k}{n^2}}{2^{k+1} + \frac{2^kk}{n^2}}.$$Making the denominator smaller makes the whole fraction bigger, so you have the comparison$$a_n\leq \sum_{k=1}^n\frac{2+\frac{2^k}{n} + \frac{k}{n^2}}{2^{k+1}}.$$Writing everything over a common ... 2 break it up. a_n=\sum \limits_{k=1}^{n} \frac{2^kn}{2^{k+1}n^2+2^kk} + \sum\limits_{k=1}^{n} \frac{2n^2}{2^{k+1}n^2+2^kk}+\sum\limits_{k=1}^{n} \frac{k}{2^{k+1}n^2+2^kk}. Since every element of each of those series is greater than zero, the series for a_n converges iff all three of the above converge, and diverges if any of the above diverge. \sum ... 0 Look at the asymptotics as x\rightarrow \infty on the first integral $$\frac{x\sin(\ln(x))}{x^2+\cos(x)} \sim \frac{x\sin(\ln(x))}{x^2} = \frac{\sin(\ln(x))}{x}$$ If we change variables $$\int_{N}^{\infty} \frac{\sin(\ln(x))}{x} dx = \int_{\ln(N)}^{\infty} \sin(u) du$$ For the second integral, use ... 0 Your function satisfies the Dirichlet conditions and the periodic extension of f over the real line is continuous at x=\dfrac{\pi}{2}, so its Fourier series converges to f(\dfrac{\pi}{2}) = 1 + \dfrac{\pi}{2} . If you wanted to check this explicitly, you would need to show that$$2\sum^{\infty}_{k=1}\dfrac{\left(-1\right)^{3k+1}}{2k-1} = ...

1

Ciao Giuseppe. The answer is negative, as @DavidC.Ullrich has already pointed out. I can provide an explicit counterexample. Consider the function $q:\mathbb{Q}\backslash\{0\}\to\mathbb{N}$ associating to $x$ the unique natural number $q=q(x)$ such that we can write $x=\tfrac{p}{q}$ an irreducible fraction. Then define $$f(x) = \cases{q(x)&if ... 3 The answer is negative. In fact if we define the Baire class B_\alpha for countable ordinals \alpha by saying B_0=C(I), B_{\alpha+1} is the set of pointwise limits of sequences in B_\alpha, and B_\alpha=\bigcup_{\beta<\alpha}B_\beta for limit ordinals \alpha then all the B_\alpha are distinct. Or so I've read; don't ask me to prove it. ... 1$$ \frac{1}{3n+1}>\frac{1}{3(n+1)} $$\displaystyle \frac13 \lim_{N \to \infty}\sum_{n = 0}^N \frac{1}{(n+1)}→∞ , therefore that subsequence is diverge. 0 Your example is not entirely clear, but I assume this is what you meant by "limit comparison":$$ \sum_{n=0}^{N} \frac{1}{3n + 1} \leq 1 + \sum_{n=1}^{N} \frac{1}{3n} = 1 + \frac{1}{3}\sum_{n=1}^{N} \frac{1}{n} $$and so$$ \lim_{N \to \infty} \sum_{n=1}^{N} \frac{1}{n} \rightarrow \infty \Rightarrow \lim_{N \to \infty} 1 + \frac{1}{3}\sum_{n=1}^{N} ...

0

You should keep in mind that convergence of the sequence $\sum_{n = 0}^N \frac{1}{3n+1}$ is, by definition, equivalent to convergence of the series $\sum_{n=0}^\infty \frac{1}{3n+1}$, as you can verify by looking up the definition of convergence of series in any textbook. So when you apply the limit comparison test for series to conclude that the series ...

0

Look for maximum of $e^{2xn-x^2}$ then use that $g<1$ to get everything less then $h(n)e^{-n^2}$ for some $h$. I can tell you more if you want.

2

The two series $$f(z)=\sum \frac{z^n}{n},\quad g(z)=\sum (-1)^n\frac{z^n}{n}$$ are a counterexample to your conjecture, because $f(z)$ converges everywhere on the unit disk except at $z=1$ while $g(z)$ converges everywhere on the unit disk except at $z=-1$ (note, indeed, that $f(-z)=g(z)$). The answer to your second question is affirmative, as one can ...

2

One may consider, $\dfrac1{x_n^2}=\pi n$, with $n=1,2,3,\ldots$, giving $$\lim _{x_n\to 0}\:\frac{\sin\left(\frac{1}{x_n^2}\right)}{x_n^2}=\lim _{n\to +\infty}\pi n \times \sin(\pi n)=0$$ and one may consider, $\dfrac1{y_n^2}=(4n+1)\dfrac{\pi}2$, with $n=0,1,2,\ldots$, giving $$\lim _{y_n\to ... 0 Make the substitution n = \frac{1}{X^2}. Then it is clear that n \to \infty \iff x \to 0 . HEnce, we have$$ \lim_{n \to \infty} n \sin n $$Now, consider subsequences multiples of pi and pi/2 and see that they converge to different limits... 1 Let's write t=\frac{1}{x^2}, such that when x\rightarrow 0, then t \rightarrow \infty. Let's notice that \lim_{x\rightarrow 0} \frac{\sin(\frac{1}{x^2})}{x^2}=\lim_{t\rightarrow \infty}t\sin(t), which doesn't exist and therefore the original limit doesn't exist. 2 By the triangle inequality,$$\int_{B_n} |f_n-f| \mathop{d\mu} \le \int_{B_n} (|f_n|+|f|) \mathop{d\mu} \le 2M \int_{B_n} \mathop{d\mu}$$where the last inequality is due to |f_n| \le M and |f| \le M. 3 use that$$\sum_{k=1}^n k^{-1/2} = \int_{1-\epsilon}^{n+\epsilon} \left(\sum_{k=1}^n \delta(x-k)\right) x^{-1/2} dx = n^{1/2} + \frac{1}{2}\int_1^n \lfloor x \rfloor x^{-3/2} dx$$(some sort of integration by parts, see Abel's summation formula ) while$$ n^{1/2} = 1+\frac{1}{2}\int_1^n x^{-1/2} dx$$hence$$A_n = 2 n^{1/2}-\sum_{k=1}^n k^{-1/2} = 1 ...

4

May be, we could use generalized harmonic numbers since $$\sum_{n=1}^x {1\over \sqrt{n}}=H_x^{\left(\frac{1}{2}\right)}$$ and use the asymptotics for large values of $n$ $$H_x^{\left(\frac{1}{2}\right)}=2 \sqrt{x}+\zeta \left(\frac{1}{2}\right)+\frac 1 {2\sqrt x}+O\left(\frac{1}{x^{3/2}}\right)$$

3

Too lazy to whip up anything fancy, but program z2 use ISO_FORTRAN_ENV,only:wp=>REAL128 real(wp) zeta,x integer k, M M = 100000000 zeta = sum([(1/sqrt(real(k,wp)),k=1,M)])- & 2*sqrt(M+0.5_wp)-1/(48*sqrt(M+0.5_wp)**3) write(*,*) zeta end program z2 Spits out $-1.46035450880958681288949915247298$, which has $29$ digits of ...

2

Note that $\cos kz = \mathrm{Re}(\cos kz + i \sin kz) = \mathrm{Re}(e^{ikz})$. Thus: $$\sum_{k \geq 1} e^{-tk} \cos kz = \sum_{k \geq 1} \mathrm{Re}(e^{-tk})\mathrm{Re}(e^{ikz}) = \sum_{k \geq 1} \mathrm{Re}(e^{k(-t + iz)}) = \mathrm{Re}\left(\sum_{k \geq 1} \left(e^{-t + iz}\right)^k\right)$$ which is a geometric series.

1

Your error resides in the fact that, unlike what you were expecting, in reality $\textrm e ^{| \ln x |} \ne | \textrm e ^{\ln x} |$. To better understand why, choose $x = \frac 1 {\textrm e}$ and convince yourself that $\textrm e ^1 \ne |\textrm e ^{-1}| = \frac 1 {\textrm e}$. To answer your second question, notice that your series is not a true power ...

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