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3 votes
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Convergence of the sequence $x_{n} = \int_{1}^{n}\frac{\cos t}{t^{2}}$ as n tends to infinity.

$|x_n-x_m| \le \left|\int_n^{m} \frac 1 {t^{2}}dt\right| \to 0$ so $(x_n)$ is Cauchy.
geetha290krm's user avatar
  • 37.5k
1 vote

Convergence of $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} a_n^3$ for $a_n = \cos(2nπ/3)/ln(1+ n)$

Use the identity: $$\cos^3(x)=\frac{3\cos(x)+\cos(3x)}{4},$$ then $$\cos^3(2nπ/3)=\frac{1+3\cos(2nπ/3)}{4}.$$ The series $$\sum_{n=1}^{\infty}\frac{1}{4\ln^3(1+n)}$$ diverges and $$\sum_{n=1}^{\infty}\...
Riemann's user avatar
  • 7,355
2 votes

Convergence of $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} a_n^3$ for $a_n = \cos(2nπ/3)/ln(1+ n)$

$c_n = \frac{1}{ln(1+n)}$ is monotonic with $\lim_{n\to \infty} c_n = 0$. Let $b_n = \cos(\frac{2n\pi}{3})$. Then $|\sum_{n = 0}^N b_n|$ is bounded (see this, for example). $\sum_{n = 1}^{\infty} c_n ...
Sgg8's user avatar
  • 1,383
0 votes

Convergence in Probability how to solve

Is this acceptable? $P(|X_m-X|> \epsilon) = P(|X_M|> \epsilon) = P(|X_m-m+m| > \epsilon) \leq P(|X_m-m| + |m|> \epsilon) = P(|X_m-E[X_m]| > \epsilon - |m|)\leq \frac{Var(X_m)}{\epsilon ^...
adisnjo's user avatar
  • 205
2 votes
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Convergence in Probability how to solve

You want to show that $\lim_{m\rightarrow\infty}P(|X_{m} - X| > \epsilon) = 0$, where $X$ is the zero random variable. Given that $X_{m}$ can take positive value $m^2$ with probability $\frac{1}{m}$...
rcescon's user avatar
  • 239
1 vote

Divergence of $\big(\frac{5k^2}{k+1}\big)^k\cdot\frac{1}{5(k+1)}$ to $\infty$

Alternatively: Let $k \ge 4$, then $x_k = 5^k\cdot \left(k-1+\dfrac{1}{k+1}\right)^k\cdot \dfrac{1}{5(k+1)}=5^{k-1}(k-1)^k\cdot\left(1+\dfrac{1}{k^2-1}\right)^k\cdot \dfrac{1}{k+1}> 5^{k-1}\cdot \...
Wang YeFei's user avatar
  • 6,783
4 votes
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Divergence of $\big(\frac{5k^2}{k+1}\big)^k\cdot\frac{1}{5(k+1)}$ to $\infty$

You can use $k+1 \le 2k$ to first get the estimate $$x_k = \left ( \frac{5k^2}{k+1}\right )^k \frac{1}{5(k+1)}\ge \left ( \frac{5k^2}{2k}\right )^k \frac{1}{10k}=\frac{1}{10} \left (\frac 52\right )^k ...
Lukas's user avatar
  • 2,511
2 votes

Divergence of $\big(\frac{5k^2}{k+1}\big)^k\cdot\frac{1}{5(k+1)}$ to $\infty$

Hint: $$\frac{5k^2}{k+1}>5(k-1),$$ and so $$x_k>\frac{5^k(k-1)^k}{5(k+1)}.\tag1$$ There are a lot of ways to proceed from here. For example, you can prove by induction that $5^{k-1}>k+1$ for $...
Thomas Andrews's user avatar
2 votes

Divergence of $\big(\frac{5k^2}{k+1}\big)^k\cdot\frac{1}{5(k+1)}$ to $\infty$

For $k\geq 2$, it is obvious that $5k^2\geq k^2\geq k+1$, so $$\left(\dfrac{5k^2}{k+1}\right)^k\geq \left(\dfrac{5k^2}{k+1}\right)^2=\dfrac{25k^4}{(k+1)^2} $$ From where $$\left(\dfrac{5k^2}{k+1}\...
Julio Puerta's user avatar
  • 5,241
2 votes

Is there an analytic continuation of the Legendre Chi function $\chi_2(z)$ for $z > 1$?

By the harmonic series $$ \chi_2'(1) = \chi_2'(-1) = \sum_{k=0}^{\infty}\frac{1}{2k+1} = \infty, $$ there are vertical asymptotes on the radius of convergence, so the answer seems to be no. But there ...
Nolord's user avatar
  • 416
2 votes

Is there an analytic continuation of the Legendre Chi function $\chi_2(z)$ for $z > 1$?

$$\begin{aligned} \chi_2(z)=\sum_{k\geq 0} \frac{z^{2k+1}}{(2k+1)^2}&=\sum_{k\geq 0}\int_0^z \frac{y^{2k}}{(2k+1)}\,\mathrm dy\\ &=\sum_{k\geq 0}\int_0^z \frac{1}{y}\int_0^yx^{2k}\,\mathrm dx\,...
Conreu's user avatar
  • 1,918
0 votes

Proving ${u_k}\to u$ given $\lim_{k\to\infty}\langle u_k,v\rangle=\langle u,v\rangle$ for $u\in \mathbb{R}^n,\forall v\in \mathbb{R}^n$

I will try to be rigoursly, and use the notation of your book and the hint. Let $u_{k},u,v \in \mathbb{R}^{n}$, such that $u_{k} \to u$, and $\{ e_{1}, \dots , e_{n} \} $ a orthonormal base of $\...
peE_'s user avatar
  • 39
2 votes
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Sufficient Condition on Almost Surely Convergence

No, it isn't enough. Let $\mathbb{P}$ be Lebesgue measure on the unit interval and $$ f_n = 1 - \chi \left( \frac{n-2^k}{2^k}, \frac{n-2^k + 1}{2^k} \right)$$ for $k\geq 0$ and $2^k \geq n < 2^{k+1}...
Jose Avilez's user avatar
-1 votes

Is 1/3 included in the sequence 0.3, 0.33, 0.333,...?

As others have pointed out, this is a matter of definition. We might translate the statement that 1/3 is included in the sequence to something a little more precise, such as: There exists an index ...
MCLooyverse's user avatar
1 vote

Convergence in Hausdorff metric (compact space)

No. In $X=[0,2]^2$, call $K=[0,1]^2$ and take an enumeration $q_n$ of $\Bbb Q^2\cap ([1,2]\times[0,1])$, starting with $q_0=(1,0)$. Define $M_n= K\cup \bigcup_{j=1}^n \{tq_{n-1}+(1-t)q_n\,:\, t\in [0,...
Sassatelli Giulio's user avatar
0 votes

Prove via contraposition that if $(a_n)$ is decreasing and $\sum_{n=1}^\infty a_n<\infty$, then $\lim_{n\to\infty}na_n=0$

This answer is based on https://math.stackexchange.com/a/4899007/1311533 I am just rephrasing in my own words. It is $\lim_{n\to\infty}n\cdot a_n\neq0$, which means $$\exists \varepsilon>0 \; \...
Olivier04's user avatar
1 vote

for which p is the infinite series of sin(1/n^p) convergent?

okay I figured myself: since $\frac{1}{n^p}$ tends to 0 we can rewrite $ \sum_{n=1}^{\infty} \sin\frac{1}{n^p}$ as: $$ \sum_{n=1}^{\infty} \frac{1}{n^p}×\frac{\sin\frac{1}{n^p}}{\frac{1}{n^p}} $$ and ...
Delta Account's user avatar
3 votes
Accepted

Prove via contraposition that if $(a_n)$ is decreasing and $\sum_{n=1}^\infty a_n<\infty$, then $\lim_{n\to\infty}na_n=0$

Suppose there exists $\varepsilon>0$ and $\sigma:\mathbb{N}\longrightarrow\mathbb{N}$ strictly increasing such that $\sigma(n)a_{\sigma(n)}\geqslant\varepsilon$ for all $n$. Since $(a_n)$ is ...
Tuvasbien's user avatar
  • 8,977
34 votes
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Is 1/3 included in the sequence 0.3, 0.33, 0.333,...?

One thing that hasn't been said in the previous answers, and really needs to be emphasized, is that mathematics is not about guesswork, so we really cannot think "so it seems ok to ...". In ...
user21820's user avatar
  • 58k
0 votes

Pointwise convergence of sequence of continuous functions to a continuous function

One more example to show that under no additional assumptions, $f_n$ need not converge to $f$ at all $x$. Define $$f_n(x) = \cos^{2n}(n!x\pi).$$ Let $f(x) = 1$. Then, at each rational number $x$, $f_n(...
Yathi's user avatar
  • 2,152
0 votes

Pointwise convergence of sequence of continuous functions to a continuous function

Unfornunately without further assumptions like moniticity, or uniform convergence on $\mathbb{Q}$ (not sure for the latter) you can't obtain the result. Consider the following sequence of continuous ...
julio_es_sui_glace's user avatar
2 votes
Accepted

Is sequence $\sum_{k=1}^{n}\sin\left( k+\frac{1}{k}\right)$ bounded? If so, does $\sum_{k=1}^{\infty}\sin k-\sin\left(k+\frac{1}{k}\right)$ converge?

First of all, your analysis seems correct: you can't use the same exact methods. If you study $u_k = \sin(k) - \sin\left(k + 1/k\right)$, you easily get $u_k \sim -\sin'(k)/k = -\cos(k)/k$. You could ...
julio_es_sui_glace's user avatar
8 votes

Is 1/3 included in the sequence 0.3, 0.33, 0.333,...?

A single term of your series can be expressed as $$A_n=\frac{1}{3}\bigg(1-\frac{1}{10^n}\bigg)$$ Using basic limits, this means that if $n→∞,A_n→\frac{1}{3}$. But according to your definition, $n$ is ...
Gwen's user avatar
  • 1,565
22 votes

Is 1/3 included in the sequence 0.3, 0.33, 0.333,...?

That is a series which converges to $\frac{1}{3}$ which, informally, means it will eventually get arbitrarily close. However, no entry in the series will be exactly $\frac{1}{3}$. Similarly, the ...
badjohn's user avatar
  • 8,474
2 votes
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Proving Existence of Sequence and Series Satisfying Hardy-Littlewood Convergence Condition Prove that for every $\vartheta, 0 < \vartheta < 1$,

For each natural $n$ put $\lambda_n=2^{n-1}$ and $a_n=(-1)^{n-1}$. Thus Conditions i and iii are satisfied. Condition ii is satisfied too. Indeed, for each $r\in (0,1)$ put $f(r)=\sum_{n=1}^{\infty} ...
Alex Ravsky's user avatar
  • 91.3k
3 votes
Accepted

how to evaluate $\sum_{n=1}^\infty \sum_{m=1}^{n-1} \frac{(-t)^{m+n}}{m-n}$?

$$ \sum_{n=1}^\infty \sum_{m=1}^{n-1} \frac{(-t)^{m+n}}{m-n} = -\sum_{n=1}^\infty \sum_{m=1}^{n-1} \frac{(-t)^{m+n}}{n-m} $$ $$ = -\sum_{n=2}^\infty \sum_{m=1}^{n-1} \frac{(-t)^{m+n}}{n-m} $$ ...
R. J. Mathar's user avatar
  • 2,539
1 vote

Limit of $\mathbb{P}(A_n|B_n)$ for $A_n,B_n\rightarrow A,B$

I suppose $1_{A_n} \to 1_A$ means $1_{A_n} \to 1_A$ almost surely. You cannot write convergence of random variable without specifying the type of convergence. This is a simple application of ...
geetha290krm's user avatar
  • 37.5k
1 vote
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Limit of $\mathbb{P}(A_n|B_n)$ for $A_n,B_n\rightarrow A,B$

Consider some sequence $(S_n)_{n\in\mathbb N}$ of sets converging to a set $S$. Let $\epsilon>0$. Then, since $S_n\rightarrow S$, we have that there is an $N\in\mathbb N$ such that $$\vert 1_{S_n} -...
Syd Amerikaner's user avatar
0 votes

Investigate convergence of $\sum_{n=1}^\infty \frac{\ln(n)}{n}$

Just use Integral test, We have $$\sum_{n \geq 1} \frac{\ln(n)}{n}$$ We know that it converges iff integral $$\int_1^\infty \frac{\ln(x)}{x} \, \mathrm{d}x$$ converges. But, \begin{align*} I &= \...
Lucky Chouhan's user avatar
0 votes

Finding a sequence for two different series so that both converge

Take $a_n=(-1)^{n}n^{-1/2}$ the first series converges by Alternating Series Test and the second one converges absolutely.
geetha290krm's user avatar
  • 37.5k
1 vote
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Sequences of nested triangles

Let $a_n,b_n,c_n$ the vertices of the $n$th triangle (think to them as complex numbers). One can express the vertices of the $n+1$th triangle in the following matrix-vector way : $$\underbrace{\...
Jean Marie's user avatar
  • 82.1k
1 vote
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Convergence Pointwise of a sequence of functions

You can just say $$\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}\frac1{n^2}\cdot\frac{x^2}{\frac{x^2}{n^2}+\left(\frac1n-x\right)^2}$$ The first term in the product goes to $0$ as $n\to\infty$, and the ...
ultralegend5385's user avatar
1 vote

Convergence Pointwise of a sequence of functions

Let $a \in (0,1).$ Then $f_n(a)= \frac{a^2}{a^2+(1-na)^2}.$ Here, see that $a$ is fixed. So, $f_n(a)$ is a sequence of real numbers. One more thing to observe is that the sequence $f_n(a) \sim \frac{1}...
MathRookie2204's user avatar
-1 votes

Can the order of convergence be defined for a bisection method?

Bisection is a simple and wasteful method. All the better methods use the available data about function values and build a model from them. The root of the model is then taken as the new midpoint. The ...
Lutz Lehmann's user avatar
0 votes
Accepted

Proving a sequence of functions is Cauchy, but not convergent in a normed space.

Your proof that $(f_n)$ is Cauchy for $\|~\|_1$ is not correct. You checked (with sometimes slightly clumsy formulations) that $$\lim_{n\to\infty}\|f_{2n}-f_n\|_1=0,$$ but this is not sufficient. What ...
Anne Bauval's user avatar
  • 35.8k
1 vote

Is $\sum_{n=1}^\infty \frac{\sin(\sqrt{n^2+1}-n)}{n }$ absolute convergent?

Another possible approach: \begin{align*} \left|\frac{\sin\left(\sqrt{n^{2} + 1} - n\right)}{n}\right| & \leq \frac{\sqrt{n^{2} + 1} - n}{n} = \frac{1}{n\left(\sqrt{n^{2} + 1} + n\right)} \leq \...
Átila Correia's user avatar
2 votes

Estimate the values of $\sum_{n\geq 1}\frac{\cos{nx}}{n}$ for $x\in\mathbb{R}$

Let $S_n(x)=\sum_{k=1}^n\frac{\cos(kx)}k$, then $$ \frac d{dx}S_{n}(x)=-\sum_{k=1}^n \sin(kx)=-\mathrm{Im}(1+e^{ix}+\dots+e^{inx}) =-\mathrm{Im}\left(\frac{1-e^{i(n+1)x}}{1-e^{ix}}\right). $$ Next, $$ ...
van der Wolf's user avatar
  • 2,380
3 votes
Accepted

Estimate the values of $\sum_{n\geq 1}\frac{\cos{nx}}{n}$ for $x\in\mathbb{R}$

From the series expansion $$ \log \frac{1}{1-z} = \sum_{n \geq 1} \frac{z^n}{n}, \quad \mbox{for}~ |z| < 1, z \in \mathbb{C}, $$ Treating the series expansion above as formal and evaluating at $z ...
Drew Brady's user avatar
  • 3,690
1 vote
Accepted

Cross product limit with angular singularity

I don't think the statement is true. Take $\Pi = \Omega + \lambda u$ and let $\lambda$ go to $0$. You have \begin{align} \frac{(\Omega \times \Pi ) \times \Pi \cdot v}{\|\Pi \times \Omega\|^2} &...
Vincenzo Tibullo's user avatar
3 votes
Accepted

Convergence of gaussian filtered function

Let $f*K_\delta=\frac{I(\delta)}{\delta\sqrt{2\pi }}$ so that $$\tag{1} I(\delta)=\int\limits_0^\infty \frac{dx'}{\sqrt{x'}} e^{-(x-x')^2/2\delta^2} $$ Changing variables $u=x-x'$ fixes the maxima of ...
Sal's user avatar
  • 4,897
1 vote
Accepted

Absolute and conditional convergence of a non-alternating series

Hint: Note that $\frac{(-1)^n}{\sqrt{n}}-\frac{(-1)^n}{\sqrt{n}+(-1)^n}=\frac{1}{n+\sqrt{n}(-1)^n}$. The original series converges if and only if the series of the term on the right converges. (why?) ...
Mark's user avatar
  • 40.4k
0 votes

Absolute and conditional convergence of a non-alternating series

$\frac{(-1)^n}{\sqrt{n}+(-1)^n}=\frac{(-1)^n}{\sqrt{n}}\frac{1}{1+\frac{(-1)^n}{\sqrt{n}}}= \frac{(-1)^n}{\sqrt{n}}(1+\frac{(-1)^n}{\sqrt{n}}+O(\frac{1}{n^2}))$ From this you can continue by using ...
aalexx.tez's user avatar
5 votes
Accepted

Convergence of square root of n series

Hint: $\frac{1}{n^{\frac{n+1}{n}}}=\frac{1}{n}\cdot\frac{1}{\sqrt[n]{n}}$. Now use the limit comparison test with a well known series.
Mark's user avatar
  • 40.4k
2 votes

Convergence of series from inverse of Cauchy product

It's natural to consider the associated entire series: $$ \begin{align} A(z) &= \sum_{n=0}^\infty a_nz^n & B(z) &= \sum_{n=0}^\infty b_nz^n & C(z) &= \sum_{n=0}^\infty c_nz^n \end{...
LPZ's user avatar
  • 3,038
3 votes

Check convergence of a trigonometric series

For your curiosity. If $$a_n=\sin \left(\frac{1}{\sqrt{n}}\right) \tan \left(\frac{1}{\sqrt{n}}\right)$$ by Taylor $$a_n=\color{red}{\frac{1}{n}}+\frac{1}{6 n^2}+\frac{31}{360 n^3}+O\left(\frac{1}{n^4}...
Claude Leibovici's user avatar
2 votes
Accepted

Interchange limit and supremum of functionals over a convex set

As requested, here is an elaboration of my comment, promoted to an answer: Assuming $B$ is bounded and closed. Since $B$ is convex, by Hahn-Banach, it is also weakly closed and therefore weakly ...
David Gao's user avatar
  • 5,205
1 vote

Check convergence of a trigonometric series

Observe that for large enough $n$, $a_n \ge \dfrac{1}{2}\cdot \dfrac{1}{\sqrt{n}}\cdot \dfrac{1}{\sqrt{n}}= \dfrac{1}{2n}$. Since $\displaystyle \sum_{n=1}^\infty \dfrac{1}{2n}$ diverges, $\...
Wang YeFei's user avatar
  • 6,783
4 votes
Accepted

Check convergence of a trigonometric series

That's a series of numbers greater than $0$, and so the comparison test can be used. You have\begin{align}\lim_{n\to\infty}\dfrac{\sin\left(\dfrac1{\sqrt n}\right)\tan\left(\dfrac1{\sqrt n}\right)}{\...
José Carlos Santos's user avatar

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