Tag Info

New answers tagged

0

1). Since $$ \int_{n\pi}^{(n+1)\pi}\left|\sin{x}\right|dx=2 $$ there is \begin{align} \int_{\pi}^{(n+1)\pi}\left|\dfrac{\sin{x}}{x}\right|dx&=\sum_{k=1}^n\int_{k\pi}^{(k+1)\pi}\left|\dfrac{\sin{x}}{x}\right|dx \\ &\geqslant\sum_{k=1}^n\dfrac1{(k+1)\pi}\int_{k\pi}^{(k+1)\pi}\left|\sin{x}\right|dx \\ &=\dfrac{2}{\pi}\sum_{k=1}^n\dfrac1{k+1} ...


0

HINTS: Fot the first part, write $$\begin{align} \int_{\pi}^{n\pi}\left|\frac{\sin x}{x}\right|dx&=\sum_{k=1}^{n-1}\int_{k\pi}^{(k+1)\pi}\left|\frac{\sin x}{x}\right|dx\\\\ &\ge \sum_{k=1}^{n-1}\frac{1}{(k+1)\pi}\int_{k\pi}^{(k+1)\pi}|\sin x|\,dx\\\\ &=\frac{2}{ \pi}\sum_{k=1}^{n-1}\frac{1}{k+1} \end{align}$$ For the second part, note that ...


1

A heuristic is that an integer $n$ is prime with "probability" one in $\ln n$, and so we can estimate the sum with its "expected" value: $$ \sum_{\substack{p \leq n \\ p \text{ prime}}} \frac{1}{p} \approx \sum_{k=2}^n \frac{1}{k \ln k} \approx \int_2^n \frac{\mathrm{d}x}{x \ln x} \approx \ln \ln n$$ In fact, the Meissel-Mertens constant is given by $$ M ...


2

By Sum of reciprocal prime numbers, $$\sum_{p \le n}{\frac1{p}} = C + \ln\ln n + O\left(\frac1{\ln n}\right)$$ Therefore $\ln \ln n$ fits the bill.


5

The easiest way to see that $$ -\log n+\sum_{k=1}^{n}\frac{1}{k} $$ is convergent is to write it as $$ -\log\left(1+\frac{1}{n}\right)+\sum_{k=1}^{n}\left(\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right)$$ and check that $\frac{1}{k}-\log\left(1+\frac{1}{k}\right)=O\left(\frac{1}{k^2}\right)$. In the same way, $$ ...


7

Yes there is a constant associated with the sum of the reciprocals of the primes. In particular, Mertens showed that $$\sum_{p \text{ prime } \le x} \frac1p - \log\log(x)$$ converges to a constant as $x\to \infty$. This is a result from 1874. I found the result in a paper: EULER’S CONSTANT: EULER’S WORK AND MODERN DEVELOPMENTS - By JEFFREY C. LAGARIAS ...


4

Bounded sequences I suppose never - because 1. To elaborate: Let $\let\epsilon\varepsilon(a_n)$ be bounded, say $|a_n|<M$. Then for $a=0$ we can let $\epsilon=M$ and see that this sequence "converges" to $0$. Or for arbitrary $a$, take $\epsilon = |a|+M$ and see that the sequence also "converges" to $a$. For the other direction assume $a_n$ "converges" ...


-1

1) You can't exchange quantifiers and you can't swap quantifications if there is a $\exists$. This may alter the meaning of the formula. 2) The sequence $a_n$ you defined does not converge in the usual sense. 3) Convergence to limit $a$ means: For any interval $(a-\epsilon, a+\epsilon)$ eventually $a_n$ stays inside that interval.


1

We will show the sequence is Cauchy. Perhaps the steps in this argument will help you understand some of the results others stated without support. As in the case of more straightforward limit proofs, we first do some scratch work. So let $m, n \in \mathbb{N}$ be given, and assume $m = n + k$ for some $k > 0$. We have to bound the difference $|x_{n+k} - ...


2

Note that $$ \lim_{n\to\infty}\sum_{k=n}^\infty\frac1{k^2}=0 $$ and for $m\gt n$, $$ a_m\le a_n+\sum_{k=n}^\infty\frac1{k^2} $$ First, take the $\limsup\limits_{m\to\infty}$: $$ \limsup_{m\to\infty}a_m\le a_n+\sum_{k=n}^\infty\frac1{k^2} $$ which must be non-negative. Then take the $\liminf\limits_{n\to\infty}$: $$ ...


4

Because the series $\sum_{n=1}^\infty\dfrac1{n^2}$ converges, to each $\varepsilon>0$ there is an $N(\varepsilon)$ such that $\sum_{n>N}n^{-2}<\varepsilon$. If $A,B$ were two distinct accumulation points, say with $A<B$, let $\varepsilon=(B-A)/3$. After having visited an $\varepsilon$-neighborhood of $A$ after $n>N(\varepsilon)$ you can never ...


1

Hint: Since $(x_n)$ is bounded, there is a convergent subsequence $x_{n_k}\to y\ge0$. Show that $\forall \epsilon>0$, $\exists N>0$, $\forall n>N$, $|x_n-y|<\epsilon$. (Note that $x_n$ is between some $x_{\displaystyle{n_k}}$ and $x_{\displaystyle{n_{k+1}}}$ , and that $\sum_{n=1}^\infty\dfrac{1}{n^2}<\infty.$)


6

For $m \gt n \gt 1$ you have $x_m-x_n \le \dfrac{1}{n-1}$ so any two accumulation points must be arbitrarily close (use $m$ for a subsequence approaching the higher accumulation point and $n$ for a subsequence approaching the lower accumulation point) and so any accumulation points must be equal.


0

If you want to work in a classical pointwise way, the inversion is a Cauchy Principle Value integral $$ f(x)=\lim_{R\rightarrow\infty}\int_{-R}^{R}\hat{f}(\xi)e^{2\pi i\xi x}d\xi . $$ If $f$ is absolutely integrable and satisfies some condition such as having left- and right-hand derivatives at $x$, or is of bounded variation on an interval $[a,b]$ with ...


1

$$G_{S_n}(z)=\left[G_{X_{n1}} (z)\right]^n=\left[\mathbb{E}z^{X_{n1}}\right]^n=\left[\frac{z}{n}+\left(\frac{z}{n}\right)^2+\left(1-\frac{1}{n}-\frac{1}{n^2}\right)\right]^n$$ $$\lim_{n\rightarrow \infty}G_{S_n}(z)=e^{z-1}$$ which corresponds to the probability generating function of $\text{Poisson(1)}$. Note: For large $n$ ...


1

On the integration range $\tan x$ is negative, hence I assume that $x^{1/3}$ is defined as $-(-x)^{1/3}$ for negative $x$. Given that, by setting $x=\pi-z$, then $z=\arctan t$, we have: $$ I = -\int_{0}^{\pi/2}\tan(z)^{1/3}\,dz = -\int_{0}^{+\infty}\frac{t^{1/3}}{1+t^2}\,dt$$ and the last integral is convergent, since: $$0\leq ...


1

HINT: $$\tan x=\frac{\sin x}{\cos x}=O\left((x-\pi/2)^{-1}\right)$$ and the integral $$\int_0^1 \frac{dx}{x^a}$$ converges for $a<1$.


4

Besides @MichaelBurr 's very smart idea exposed in a comment under the original post, you could also use the change of variable $t = x^2$ which transforms your integral into $\frac 1 2 \int \limits _0 ^\infty \frac {\sin t} t \Bbb d t$, to which you can now immediately apply the Abel-Dirichlet theorem.


2

Note that $0 \leq \frac{e^{-2x}}{\sqrt{x+16}} \leq \frac{e^{-2x}}{4}$ for $x \geq 0$ now compute $$\int_1^\infty \frac{e^{-2x}}{\sqrt{x+16}} \leq \int_1^\infty\frac{e^{-2x}}{4}< \infty$$


1

Consider by itself $$ \frac{1}{\sqrt{x + 16}}$$ It is clear that for $1 \le x $ $$ 0 \le \frac{1}{\sqrt{x + 16}} \le \frac{1}{\sqrt{17}}$$ Thus we conclude that $$ \int_1^{\infty} \frac{e^{-2x}}{\sqrt{x+17}} dx \le \frac{1}{\sqrt{17}} \int_0^\infty e^{-2x} dx = \frac{1}{\sqrt{17}} \frac{1}{2} $$ Furthermore since $$ \frac{e^{-2x}}{\sqrt{x+16}} > ...


1

You don't like the term "conditionally convergent" because it's redundant -- it can be defined in terms of other things we already have -- and misleading, because it sounds as if we're saying something might be convergent when we know that it IS convergent. The first is (I think) a not very good reason to dislike a definition. Almost everywhere in ...


1

We say "conditionally convergent" if a series converges, but not absolutely, because according to the Riemann theorem we can rearrange the terms so that we sum up all the same terms (just in a different order) and yet the limit of the sum is either non-existent, or equal to $\pm \infty$ (or equal to any desired finite result, but anyways, we say ...


1

You certainly need to assume that $\hat f\in L^1$ to apply the inversion formula as written; no, that does not follow from assuming $f\in L^1$. It's been pointed out that $f.f'.f''\in L^1(\mathbb R)$ imply $\hat f\in L^1$; you might also note that $f,f'\in L^2(\mathbb R)$ imply $\hat f\in L^1$. Even if $\hat f\notin L^1$ you can still recover $f$ from $\hat ...


1

Hint: Write $$\sum \left(1 - \frac{a_n}{a_{n + 1}}\right)\frac{1}{\sqrt{a_{n + 1}}}= \sum(a_{n + 1} - a_n)\frac{1}{a_{n + 1}^{3/2}}.$$ Then try to use the Abel's test.


0

Hint Note $$\left(1-\dfrac{a_{n}}{a_{n+1}}\right)\dfrac{1}{\sqrt{a_{n+1}}}=\dfrac{a_{n+1}-a_{n}}{a_{n+1}\sqrt{a_{n+1}}}\le\int_{a_{n}}^{a_{n+1}}\dfrac{1}{x\sqrt{x}}dx$$


1

No, it doesn't imply absolute integrability. Yes, both $f$ and $\hat f$ need to be $L^1$. However, if for example $f\in C^2(\mathbb R)$ and $f',f''\in L^1$, then $$\mathcal F [f''](\xi)=(-2\pi i \xi)^2 \hat f (\xi)$$ is a bounded continuous function, and for this to be true $\hat f$ must be bounded at least by something like $$|\hat f(\xi)|\leq ...


3

The initial version of this proof was mistaken. All credit for this one should go to @grizzly , who suggested this approach in a comment below. Note that $$\lim \limits _{x \to 1^-} (1-x) \sum \limits _{k=1} ^\infty \frac {k a_k x^k} {1-x^k} = \lim \limits _{x \to 1^-} \sum \limits _{k=1} ^\infty \frac {(1-x) k a_k x^k} {(1-x) (1 + x^2 + x^3 + \cdots + ...


2

$a_{n}$ converges to $a$ so given $t>0$ there exists $N$ such that $n \geq N \Rightarrow |a_{n}-a|<t$. Let $\epsilon >0$ and $t=\frac{\epsilon}{c}$. $(c>0)$


2

Your first reason is the right one. Let $\epsilon'=\frac{\epsilon}{|c|}$, then by the existence of the original limit, there exists $M$ such that $\forall n>M,|a_n-a|<\epsilon'=\frac{\epsilon}{|c|}$


0

Over the interval $I=\left(0,\frac{\pi}{2}\right)$, $0<\sin x<x$, hence for any $\varepsilon\in I$: $$\int_{\varepsilon}^{\pi/4}\frac{dx}{x\sin(2x)}\geq\int_{\varepsilon}^{\pi/4}\frac{dx}{2x^2}=\frac{1}{2\varepsilon}-\frac{2}{\pi}$$ and your integral is clearly divergent.


0

Note that $$\sin\left(2x\right)\leq 2x $$ in all real range $\left(0,\infty\right) $, then $$\int_{0}^{\pi/4}\frac{1}{x\sin\left(2x\right)}\geq\int_{0}^{\pi/4}\frac{1}{2x^{2}}dx=\infty. $$


-2

You can bring it to $\int \frac{dx}{x\sin x}$ integral with some replacements which doesn't converge.


1

You can approximate by an integral this way: $$ \int_x^{2x} \frac{\mathrm d x} {x} = \ln(2x)-\ln x = \ln 2 \approx 0.6931 $$ You would need to prove that the difference between the integral and the summation vanishes for $n \rightarrow \infty$. One way to do this is to note that the summation is actually a Riemman sum that converges to the integral.


5

Using Riemann summation : $\displaystyle\sum_{k=1}^n\frac{1}{n+k}=\frac{1}n\sum_{k=1}^n\frac{1}{1+\frac{k}n}=_{n\infty}\int_0^1\frac{dx}{1+x}=\ln2$


1

$$\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}=\sum_{k=1}^{n}\frac{1}{n+k}=\frac 1n\sum_{k=1}^{n}\frac{n}{n+k}=\frac 1n\sum_{k=1}^{n}\frac{1}{1+\frac kn}.$$


1

Note $$1<1+x+x^2+\cdots+x^{n-1}<n,x\in(0,1)$$ so $$x^n-1<\dfrac{nx^n}{1+x+\cdots+x^{n-1}}-1<nx^n-1$$ and we know $\sum_{n=1}x^n,\sum_{n=1}^{+\infty}nx^{n}$ are converge then you know?


1

Given that $N_m=k$, where $k\ge 1$, and $0\lt x\lt 1$, we have $$\Pr(M_m\le x)=x^k.$$ Now we need to decide what $\Pr(M_m\le x)$ is if $N_m=0$. There is no obvious definition. With not much conviction, we call this probability $1$. Then $$\Pr(M_m\le x)=\sum_{k=0}^\infty e^{-m} \frac{m^k}{k!}x^k.$$ We recognize the sum as $e^{-m}e^{mx}$.


1

First, $k=1,\dots,n$ (otherwise, the total prob. does not sum up to 1). Then $$\varphi_{X_n}(t)=\mathbb{E}[e^{\mathrm{i}tX_n}]= \sum_{k=1}^{n}e^{\mathrm{i}t\frac{k}{n}}\cdot\frac{1} {n}= \frac{\exp\left(\mathrm{i}t\cdot\frac{n+1}{n}\right)-\exp\left(\mathrm{i}t\cdot\frac{1}{n}\right)}{\left(\exp\left(\mathrm{i}t\cdot\frac{1}{n}\right)-1\right)\cdot n} ...


0

I think if you use $\ln(1-x)<-x$, then you can show that $% %TCIMACRO{\dprod }% %BeginExpansion {\displaystyle\prod} %EndExpansion \left( 1-a_{i}\right) >0\rightarrow% %TCIMACRO{\dsum }% %BeginExpansion {\displaystyle\sum} %EndExpansion \ln\left( 1-a_{i}\right) >-\infty\rightarrow% %TCIMACRO{\dsum }% %BeginExpansion {\displaystyle\sum} ...


1

Split the integral into $\int_0^1 + \int_1^{\infty} $. For the first integral, the function is less than 1, so the integral is also less than 1. For the second integral, the function is less than $1/x^2$, and the integral of this from 1 to $\infty$ is 1.


0

The answers above, as usual, are well-written and correct. I just wanted to add some thoughts: If you just wanted to memorize the geometric sums formula to solve the problem, that would work. But it's sometimes more helpful (especially with formulas that are rather easily derived) to know the derivations, because it allows you to get a deeper understanding. ...


1

$$\sum_{i\ge 0}ar^i=\frac a{1-r}=12.$$ $$\sum_{i\ge 0}ar^{2i+1}=\frac{ar}{1-r^2}=\frac a{1-r}\frac r{1+r}=\frac{12r}{1+r}=5,$$ whence $\,r=\dfrac57,\quad a=\dfrac{24}7$.


0

HINT: The formula for a convergent geometric sum tells you that $$ 1 + r + r^2 + r^3 + \cdots = \frac{1}{1-r} . $$ How can you determine the value of the sum with the odd powers, which is $ r + r^3 + r^5 + r^7 + \cdots ? $ Note that $$ r + r^3 + r^5 + r^7 + \cdots = r(1 + r^2 + r^4 + r^6 + \cdots . $$


0

The sum of an infinite geometric sequence is $\frac{a}{1-r}$. So, you get the equation $\frac{a}{1-r}=12$, which is $a=12-12r$. Now, the second part is basically the geometric sequence $ar+ar^3\cdots$. The sum is $\frac{ar}{1-r^2}$, which is equal to $5$, which simplifies to $ar=5-5r^2$. Setting these two parts equal you get $12r-12r²=5-5r^2$. This is a ...


1

You've got the right idea. We want, for any choice of $\epsilon$, to be able to find an $M$ such that the inequality holds. The process you worked through (i.e. taking $M > \frac{1}{\epsilon^2}$) gives us a way to find/construct our $M$ in such a way that it works for any $\epsilon$ we like. The proof lies in being able to find an $M$ for any ...


1

You have to show, that $\forall \varepsilon>0 \ \exists M: |a_n-a|<\varepsilon \ \forall n\geq M $. So your argumentation is correct up to this point. And you are done because $\varepsilon>0$ was arbitrarily choosen and you have found an $M$. So the proof is done. You only have to show existence and that is what you have done.


1

I think your last statement is closest. We want to show that for every $\varepsilon$, it is possible to find the appropriate $M$ such that. Showing that this is possible by construction of $M$ is usually a good way to go.


8

Let $I(a)$ be the integral $$I(a)\equiv \int_0^{\infty}\log\left(1+\frac{a^2}{x^2}\right)dx$$ Taking a derivative with respect to $a$ reveals that $$\begin{align} I'(a)&=\frac{d}{da}\int_0^{\infty}\log\left(1+\frac{a^2}{x^2}\right)dx\\\\ &=2a\int_0^{\infty}\frac{dx}{x^2+a^2}\\\\ &=\pi \end{align}$$ Integrating shows that $I(a)=\pi a +C$. ...


6

Using integration by parts: $$\int_0^\infty \ln\left(1+\frac{a^2}{x^2}\right)\,dx\,\,\begin{bmatrix}u=\ln\left(1+\frac{a^2}{x^2}\right)& du=\frac{\frac{-2a^2}{x}}{x^2+a^2}\,dx \\ dv=dx& v=x\end{bmatrix}\\=\underbrace{\left.x\cdot \ln\left(1+\frac{a^2}{x^2}\right)\right|_0^{\infty}}_{L}+\underbrace{\int_0^{\infty}\frac{2a^2}{x^2+a^2}\,dx}_{I}$$ ...


5

$$\int_{0}^{+\infty}\log\left(1+\frac{a^2}{x^2}\right)\,dx = a\int_{0}^{+\infty}\log\left(1+\frac{1}{x^2}\right)\,dx $$ then by setting $x=\tan\theta$ we have: $$ \int_{0}^{+\infty}\log\left(1+\frac{1}{x^2}\right)\,dx = -2\int_{0}^{\pi/2}\frac{\log\sin\theta}{\cos^2\theta}\,d\theta=2\int_{0}^{\pi/2}\cot(\theta)\tan(\theta)\,d\theta=\color{red}{\pi},$$ where ...



Top 50 recent answers are included