# Tag Info

1

$\dfrac{1}{n!} \geq \dfrac{1}{n^n}$ and this proves the second series diverges by comparison to the harmonic series. The first one is already asked here, and I just quoted its proof: $\sqrt[n]{n} - 1 \geq c\dfrac{\ln n}{n}, c > 0$ and this means it diverges by comparison test as well since the right series diverges by again comparing it to the harmonic ...

1

This is an interesting question! I thought I would get the ball rolling by looking first at the case for the square pool (4-gon). A diagram: The teacher is located at point $A$, the boy is in the center of the pool at point $E$. The square has the size $2x2$, the speed of the boy is arbitrarily set at $1$ and the speed of the teacher is $K > 1$. ...

1

You're on the right track, but a few things need to be mentioned. First, if the series involves complex numbers then you're actually looking for the disk of convergence, not the interval. Second, your notation is very inappropriate. When you apply the root test, you're applying it to $a_n$ only. Including the summation symbol is incorrect. Third, your ...

0

$\sin x$ is an entire function, hence $$1-n\sin\frac{1}{n}=-\sum_{m\geq 1}\frac{(-1)^m}{(2m+1)!n^{2m}}\tag{1}$$ holds for every $n\in\mathbb{N}^*$ and: $$\sum_{n\geq 1}\left(1-n\sin\frac{1}{n}\right) = \color{red}{\sum_{m\geq 1}\frac{(-1)^{m+1}\zeta(2m)}{(2m+1)!}} \tag{2}$$ where the RHS of $(2)$ is a fast-converging series with alternating signs, ...

1

As can be shown using $x>\sin(x)$, the function $$\frac{x-\sin(x)}{x^3}$$ has a negative derivative for $x\ge0$. By L'Hospital, the limit at $0$ is $\dfrac16$ so that $$0\le\frac{x-\sin(x)}{x^3}\le\frac16$$ and $$0\le1-\frac{\sin(x)}x\le\frac{x^2}6.$$ Summing for all $\dfrac1n$, by the Basel problem the series converges and $$\sum_{n=1}^\infty\left(1-... 3 Note that, for large n, by Taylor expansion$$\sin\left(\frac{1}{n}\right)=\frac1n+\frac{1}{3!n^3}+o\left(\frac{1}{n^3}\right),$$hence the summands are of the form$$1-n\cdot\left(\frac1n+\frac{1}{3!n^3}+o\left(\frac{1}{n^3}\right)\right)=-\frac{1}{3!n^2}+o\left(\frac{1}{n^2}\right).$$Thus the series converges. 0 Take any series you are familiar with and has a finite radius of convergence r. Then rescale the argument to be \dfrac{3z}r. This multiplies all terms by \dfrac{3^n}{r^3} and yields the desired radius. 0 Taking the center of the desired power series at origin, the domain of convergence must be |z|<3. Indeed, you need a power series of the form f(z)=\sum_{n=0}^{\infty}$$a_n.z^n$which converges uniformly if$|z/3|<1$. One such choice is$f(z)=\frac{1}{3-z}=\sum_{n=0}^{\infty}\frac{1}{3^{n+1}}.z^n. 0 as 0<a<1 , so 1/a >1 so \lim _{x \to \infty } xa^x=\lim_{x \to \infty }x/a^{-x}=0 ; now as x_n>0 and \lim x_n=0 , so \lim \dfrac 1{x_n}=\infty , so by sequential criteria of limit at infinity , 0=\lim _{x \to \infty } xa^x=\lim_{n \to \infty} (\dfrac 1{x_n})a^{\dfrac 1{x_n}}=\lim_{ n\to \infty } \dfrac {a^{1/x_n}}{x_n} , so by ... 1 Put b=-\log a >0, f(y)=y\exp(-by). It is easy to see that f is continuous on I=[0,+\infty[, and that f(y)\to 0 if y\to +\infty. Hence f is bounded, say by M, on I. We have f(1/x_n)\leq M, hence a^{1/x_n}\leq M x_n, and it is easy to finish. 0 Assume that \sum x_n <1, then pick a_n= \inf\{k | x_m<\frac{1}{n}, \forall m\geq k \} for n\geq 1. Notice that a_n \leq n, and that a_n is increasing. \begin{align*} \sum _{n=1}^{\infty} a^{\frac{1}{x_n}} &= \sum_{k=1}^{\infty}\sum _{n=a_k}^{a_{k+1}} a^{\frac{1}{x_n}} \\ &\leq \sum_{k=1}^{\infty}\sum _{n=a_k}^{a_{k+1}} a^{n}\\ &... 0 I don't think I have enough of your problem to answer that question, but if you look at Huber's Robust Statistics or Keener's Theoretical Statistics, the sections on M-estimators might help. For example, in Keener, the maximum likelihood estimator is characterized as when the derivative of the likelihood equals 0. They are able to do proofs using a Taylor ... 0 The problem in the presented proof is that it only show that for any fixed k, \lim_{n\to +\infty}\mathbb P\left(\left|S_n-S_k\right|/n\gt\varepsilon \right)= 0, which is not hard to see from the convergence in probability of \left(S_n/n\right)_{n\geqslant 1} to 0. We actually have to prove that \lim_{n\to +\infty}\color{red}{\max_{1\leqslant k\... 1 \frac{1}{(1+x)^2} is a non-negative and bounded function on the integration range, hence the given integral is convergent since \int_{0}^{1}\log(1-x)\,dx = \int_{0}^{1}\log(x)\,dx = -1. \tag{1}$$Through the substitution x=1-e^{-t}, the original integral equals:$$ \int_{0}^{+\infty}\frac{-te^{-t}}{(2-e^{-t})^2}\,dt=-\sum_{n\geq 0}\int_{0}^{+\infty}\...

1

If you want to calculate the value of the integral (in case of convergence) anyway, you could just use the definition of the improper integral and see if the following limit exists: $$\lim_{b \to 1^-} \int_0^b \frac{\ln(1-x)}{(1+x)^2} \,\mbox{d}x$$ If this limit exists, you have not only proven convergence of the improper integral, you also have its value. ...

2

Notice that $- \frac{\ln (1-x)}{ 1+x^2} \leq -\ln (1-x)$ and prove the convergence of $$\int _0^1 \ln (1-x)d x$$

3

For a meromorphic function, the radius of convergence of the Taylor series centered at some point is just the distance from the closest singularity. $z^2+3z-4$ vanishes at $z=1$ and $z=-4$, but $z=1$ is not a singularity since $z=1$ is also a root of $z^3-1$. So the radius of convergence at the origin is just $\color{red}{\rho=4}$. Counter-proof: $$\frac{z^... 1 Let we consider:$$ \int_{0}^{+\infty}\color{red}{\frac{1}{x}}\color{blue}{\left(\sum_{n\geq 1}\frac{e^{-\pi n^2 x}-1}{n^3}-\sum_{n\geq 1}\frac{e^{-2\pi n^2 x}-1}{n^3}\right)}\,dx.$$The integral is convergent by Dirichlet's test, since the red term is decreasing toward zero while the blue term has a bounded primitive:$$\int_{0}^{t}\sum_{n\geq 1}\frac{e^{-...

2

What you have done seems Ok. It is important that $\alpha>0$ to ensure the convergence of the integral. Concerning the closed form of the integral, you may perform a change of variable $$u=\sqrt{x},\quad du=\frac1{2\sqrt{x}}dx,$$ obtaining $$I_{\alpha }=\int _0^{\infty }\left(\frac{e^{-\alpha x}}{\sqrt{x}}\right)\:dx=2\int _0^{\infty }e^{-\alpha u^2}\:... 1 Yet you can show the convergence of \sum_{n=1}^{\infty} \left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right) as follows, which is based just on the inequality \begin{gather*}\sin(x)<x,\qquad \forall x>0.\end{gather*} Indeed, since for all n\in\mathbb{N}, 1/n>0, we arrive at 0<1/n-\sin(1/n), and so the series under consideration is of ... 0 Using functional approach you can write u_n = \dfrac{2e^n}{1+e^n}=f(e^n)\implies f(x) = \dfrac{2x}{1+x}\implies f'(x) = \dfrac{2}{(1+x)^2} > 0. Thus u_n is monotonically increasing. 2 As e^{-n} > e^{-n-1} we get$$ 1 + e^{-n} > 1 + e^{-n-1} $$and thus$$ \frac{2}{1 + e^{-n}} < \frac{2}{1 + e^{-n-1}}. $$0 The "p-test" is a special case of comparison tests for integrals; it's easy to remember and very useful, but it doesn't apply to every integral. One should be willing to go back directly to the comparison tests themselves. In this case, for convergence we need a larger function whose integral still converges on [0,2]. Since x+2>2 for x\in[0,2], ... 2 Since$$\left\{ \left| \frac{S_n-S_k}{n} \right|< \epsilon \right\} \supseteq \left\{ \left| \frac{S_n}{n} \right|< \frac{\epsilon}{2} \right\} \cap \left\{ \left| \frac{S_k}{n} \right| < \frac{\epsilon}{2} \right\}$$we have$$\mathbb{P} \left( \left| \frac{S_n-S_k}{n} \right| < \epsilon \right) \geq \mathbb{P} \left( \left| \frac{S_n}{n} \...

0

If you are aware of harmonic numbers and digamma functions, then $$S_1=\sum_{n=0}^{p} \frac{1}{n+1}=\psi (p+2)+\gamma=H_{p+1}$$ $$S_2=\sum_{n=0}^{p}\frac{1}{n+z}=\psi (p+z+1)-\psi(z)$$ Now, considering the asymptotics for large values of $p$ $$S_1-S_2=(\psi (z)+\gamma )+\frac{1-z}{p}+\frac{z^2+z-2}{2 p^2}+O\left(\frac{1}{p^3}\right)$$ which make $$\sum_{... 1 Just continuing from Mark's answer, in terms of Gregory coefficients:$$\begin{eqnarray*}\text{PV}\int_{0}^{2}\frac{dx}{\log(x)} &=& \lim_{\varepsilon\to 0^+}\int_{\varepsilon}^{1}\left(\frac{1}{\log(1+x)}+\frac{1}{\log(1-x)}\right)\,dx\\&=& 2\sum_{n\geq 0}\frac{G_{2n+1}}{2n+1} \\&=&\int_{0}^{+\infty}\frac{\log(x+2)-\log(x)}{\pi^2+\...

0

If $1\ne x$ then $\log x < x-1$. You can tell that's true just because the tangent line to $y=\ln x$ at $x=1$ is $y=x-1$ and the graph is concave downward at that point. Therefore $$\int_1^2 \frac {dx} {\log x} > \int_1^2 \frac {dx} {x-1} = +\infty.$$ In a similar way, one can deduce that $$\int_0^1 \frac{dx}{\log x} = -\infty.$$ In this sort of ...

3

PRIMER: In THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the logarithm function satisfies the inequalities $$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$ for $x>0$. Note from $(1)$, that we have for $x>0$ $$\frac{x}{x-1} \ge \frac{1}{\log(x)}\ge \frac{1}{x-1}... 0 The Cauchy condensation test gives that your series is convergent iff$$ \sum_{m\geq 0}\frac{1}{1+m^2}=\frac{1+\pi\coth \pi}{2} $$is convergent. 0 \lim_{n\rightarrow \infty} \frac{p(n+1)}{p(n)}=1 for any polynomial of degree k \ge 1.Therefore radius of convergence is 1 for given power series. 0 Let us assume for some region in [0,\infty] f(x) is positive and say it P, P may contain many non-continuous sets and call remaining N for which f(x) is negative. Now, for N region in [0,\infty] we have$$|f(x)| = -f(x)= g(x)$$Than,$$\int_0^\infty|f(x)|dx = \int_Pf(x) +\int_Ng(x) $$As, \int_Pf(x) and \int_Ng(x), both are positive and sum of ... 0 Both in improper integrals and infinite series: absolute convergence \;\implies\; convergence, because with Cauchy criterion:$$\left|\int_a^bf(x)\,dx\right|\le\int_a^b|f(x)|\,dx\;\left(=\left|\int_a^b|f(x)|\,dx\right|\right)$$2 Note that \cos is 2\pi-periodic and that on [0, 2\pi], \min(0, \cos x) = 0 if x < \pi/2 or x > 3\pi/2 and \min(0, \cos x ) = \cos x otherwise. So$$b_n = \frac{1}{\pi} \int_0^{2\pi} \min(0, \cos x) \cos nx \, dx = \frac{1}{\pi} \int_{\pi / 2}^{3\pi / 2} \cos x \cos nx \, dx.$$3 Let us assume to avoid trivialities that U is required to contain no sets of cardinality smaller than |I|. Even with this assumption, no matter how large I is, your desired conclusion will not hold for arbitrary ultrafilters. Indeed, let I be any infinite set and partition I into countably many sets I_n with the same cardinality as I. Define ... 7 Note that$$\frac{1}{n\log^2(n)+n}\le \frac{1}{n\log^2(n)}$$and$$\int_2^\infty \frac{1}{x\log^2(x)}\,dx=\frac{1}{\log(2)}$$1 Should I use the derivate of it? or proving by \frac{a_{n+1}}{a_{n}} as my teacher did ? You may use standard properties of precalculus, from$$ n+1\geq n $$you get$$ (n+1)^2+(n+1)+1\geq n^2+n+1 $$giving, for n\geq1,$$ \frac1{(n+1)^2+(n+1)+1}\leq \frac1{n^2+n+1}. $$3 Convergence of X_n in L^1 does not imply convergence in L^2. For a counterexample, let$$\mathbb P\left(X_n = n^{\frac12}\right)= n^{-1} = 1-\mathbb P(X_n=0). $$Then$$\mathbb E\left[|X_n|\right] = n^{-1}n^{\frac12} = n^{-\frac12}\stackrel{n\to\infty}\longrightarrow 0, $$so that X_n\stackrel{L^1}\longrightarrow 0, but$$\mathbb E\left[\left|X_n\...

1

For a), how about $f_n(x) =\min(x,n)$? For b): Let $f(x)=\sin(x)/x$ for $x>0$ and $=0$ for $x\le 0$. And $f_n(x) =f(x)$ for $x\le n\pi$ and $=0$ for $x\ge n\pi$.

0

I had a mistake writing the formulas in the original question, now is corrected and is clear that the remainder converges to zero because the productory converges and $\left|\frac{x}{1-c_N}\right|<1$ because $$0\le |c_N|<|x|<1/2$$ Then now we have $$\lim_{n\to\infty}\mathcal R_N[f(x)]=\lim_{n\to\infty}\left(\frac{x}{1-c_N}\right)^{N+1}(1-c_N)^{-\... 1 HINT: \log(1+\frac1n) behaves as 1/n and \sum \frac{1}{n^\beta} converges for \beta>1. (Use the limit comparison test.) 1 We have$$1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}\le 1+\int_1^n \frac{dx}{\sqrt{x}}.$$Integrate and divide by n to conclude that our sequence has limit 0. Remark: Your method is correct, but requires somewhat more machinery than the above estimate. But if that machinery is already available, it provides a computation-free proof. 1 By RMS-AM,$$\sqrt{\frac{1 + 1/2 + \dots + 1/n}{n}} \geq t_n \geq 0.$$Since \displaystyle \sum_{k=1}^n \frac{1}{k} = O(\log n), we have t_n \to 0 by squeezing. 1 If the remainder term R_N does not converge to 0 then the power series will not sum to f(x). For just the case 0\leq x<1/2 :\quad Let x=1/2-y. Then c_N (which you write as c) satisfies 1-c_N>1-x for every N.  Let z=x/(1-x). Then 0\leq z<1.  For every N we have$$0<x/(1-c_N)<z<1.$$Note that z does not depend ... 1 Continuity of this relation is defined to be that if it is preserved in limits: if for any pair of sequences (x^n,y^n) converging to x and y respectively and x^n\succsim y^n \forall n, then x\succsim y. This statement easily implies for any sequence of points \{y^n\} with x\succsim y^n \forall n and y^n converging to y, we have x\... 1 The equation satisfies the conditions that guarantee existence and uniqueness of solution. x(t)=\pi/2 and x(t)=0 are constant solutions. If x(0)\in(0,\pi/2), by uniqueness of solution it must be that$$ 0<x(t)<\frac{\pi}{2} $$for all t for which thwe solution is defined. This implies that the solution is global, that is, it is defined on (-\... 1 The sequence does not converge with respect to the norm topology.The sequence is not even bounded in norm, as \|f_n\|^2=\int_{-1/n}^{1/n}n^2\;dx= 2 n. It also does not converge weakly. Otherwise \int_R f_n(x)g(x)\;dx converges for each g\in L^2(R). But this is not so. Example: Let g(x)=x^{-1/3} for x\in (0,1], and g(x)=0 for x\not \in (0,1]. ... 2 Since the integrand is a continuous function over each compact of the form [0,M], M>0, then we are left to see what happens as x \to \infty, where we have:$$ \frac{\tanh x}{1 + x^a} \sim \frac1{x^a} $$which gives a finite integral if and only if a>1, so the initial integral is convergent if and only if a>1. Since the integrand is a ... 1$$\sum_{n=0}^{\infty} \frac{6}{4n-1} - \frac{6}{4n+3}=24\sum_{n=0}^{\infty}\frac{1}{(4n-1)(4n+3)}$$Now as 4n-1>n f0r n>0 and 4n+3>n so$$\frac{1}{(4n-1)(4n+3)}\leq \frac{1}{n^2}$$2 \sum_{n=0}^m(6/(4 n-1)-6/(4 n+3)) = -6/(4 m+3)-6 where m is finite for the rearranging, but now we let it go \to \infty. What happens? 2 Dirichlet's test does the job pretty nicely. Let we prove that the partial sums of \cos n are bounded:$$ \left|\sum_{n=1}^{N}\cos(n)\right| = \left|\frac{\sin\left(N+\frac{1}{2}\right)-\sin\left(\frac{1}{2}\right)}{2\sin\left(\frac{1}{2}\right)}\right|\leq\frac{1+\sin\frac{1}{2}}{2\sin\frac{1}{2}}<\frac{14}{9}  and we are done, since \$\left\{\frac{1}{...

Top 50 recent answers are included