Question

Does $\frac1n\sum\limits_{k=1}^na_k^2\to\rho$ with $0\le\rho<1$ imply$\prod\limits_{k=1}^na_k\to0$?

score 70 · Answer 1 · 2016-05-31 06:11:31

70

Views

Does $\frac1n\sum\limits_{k=1}^na_k^2\to\rho$ with $0\le\rho<1$ imply$\prod\limits_{k=1}^na_k\to0$?

Published on 31 May 2016 - 6:11

score 85 · Answer 2 · 2016-06-01 18:40:26

85

Views

Uniformly convergence on a set implies a smaller set

Published on 01 Jun 2016 - 18:40

score 245 · Answer 3 · 2016-06-03 14:02:04

245

Views

Computing an infinite product $\prod_{n=1}^{\infty} \frac{1}{2}(1+\cos\frac{x}{2^n})$

Published on 03 Jun 2016 - 14:02

score 410 · Answer 4 · 2016-06-05 02:18:18

410

Views

Principal argument summation

Published on 05 Jun 2016 - 2:18

score 872 · Answer 5 · 2026-03-30 01:51:37

872

Views

What are practical applications of infinite products?

Published on 30 Mar 2026 - 1:51

score 55 · Answer 6 · 2016-06-05 08:53:49

55

Views

Algebraic simplifications

Published on 05 Jun 2016 - 8:53

score 237 · Answer 7 · 2016-06-08 19:25:12

237

Views

Prove ${\Gamma^2(1/4)\over 4\sqrt{2\pi}}=\prod_{n=1}^{\infty}\left({2n+1\over 2n}\right)^{(-1)^{n+1}}$

Published on 08 Jun 2016 - 19:25

score 275 · Answer 8 · 2016-06-10 15:32:18

275

Views

Showing $\int_{0}^{1}{x^{2n}-x\over 1+x}\cdot{dx\over \ln{x}}=\ln\left({2\over \pi}\cdot{(2n)!!\over (2n-1)!!}\right)$

Published on 10 Jun 2016 - 15:32

score 495 · Answer 9 · 2016-06-17 07:20:00

495

Views

Closed form of infinite product $\prod\limits_{k=0}^\infty 2 \left(1-\frac{x^{1/2^{k+1}}}{1+x^{1/2^{k}}} \right)$

Published on 17 Jun 2016 - 7:20

score 1.2k · Answer 10 · 2016-06-19 10:47:40

1.2k

Views

Infinite product of sinc function

Published on 19 Jun 2016 - 10:47

score 244 · Answer 11 · 2016-06-19 18:37:24

244

Views

An infinite product for $\arccos \alpha$ similar to Viete's formula for $\pi$

Published on 19 Jun 2016 - 18:37

score 354 · Answer 12 · 2026-03-29 02:12:41

354

Views

Closed form for $\prod\limits_{l=1}^\infty \cos\frac{x}{3^l}$

Published on 29 Mar 2026 - 2:12

score 78 · Answer 13 · 2016-06-28 23:10:37

78

Views

Showing that $\prod_{k=1}^\infty \cos\left(\frac{2\pi t}{3^k}\right)$ does not vanish at infinity.

Published on 28 Jun 2016 - 23:10

score 439 · Answer 14 · 2013-04-02 19:15:21

439

Views

Infinite product simplification

Published on 02 Apr 2013 - 19:15

score 954 · Answer 15 · 2013-04-03 02:51:43

954

Views

$\sum|a_n|^2<\infty$ and $\sum a_n<\infty$ implies $\sum\log(1+a_n)$ converges

Published on 03 Apr 2013 - 2:51

List Question

Does $\frac1n\sum\limits_{k=1}^na_k^2\to\rho$ with $0\le\rho<1$ imply$\prod\limits_{k=1}^na_k\to0$?

Uniformly convergence on a set implies a smaller set

Computing an infinite product $\prod_{n=1}^{\infty} \frac{1}{2}(1+\cos\frac{x}{2^n})$

Principal argument summation

What are practical applications of infinite products?

Algebraic simplifications

Prove ${\Gamma^2(1/4)\over 4\sqrt{2\pi}}=\prod_{n=1}^{\infty}\left({2n+1\over 2n}\right)^{(-1)^{n+1}}$

Showing $\int_{0}^{1}{x^{2n}-x\over 1+x}\cdot{dx\over \ln{x}}=\ln\left({2\over \pi}\cdot{(2n)!!\over (2n-1)!!}\right)$

Closed form of infinite product $\prod\limits_{k=0}^\infty 2 \left(1-\frac{x^{1/2^{k+1}}}{1+x^{1/2^{k}}} \right)$

Infinite product of sinc function

An infinite product for $\arccos \alpha$ similar to Viete's formula for $\pi$

Closed form for $\prod\limits_{l=1}^\infty \cos\frac{x}{3^l}$

Showing that $\prod_{k=1}^\infty \cos\left(\frac{2\pi t}{3^k}\right)$ does not vanish at infinity.

Infinite product simplification

$\sum|a_n|^2<\infty$ and $\sum a_n<\infty$ implies $\sum\log(1+a_n)$ converges

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