Real Analysis Real Analysis Math Forum

 November 1st, 2019, 10:15 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math Compute infinite sum Compute: $\displaystyle \sum_{k=1}^{\infty } \frac{\sin(kn\pi )}{k}\; , n\in \mathbb{N}.$ November 1st, 2019, 11:06 AM #2 Senior Member   Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 Math Focus: Algebraic Geometry The sum is $0$ since $kn$ will always be natural so $\sin{(k n \pi)}$ will always be $0$ If $kn$ is odd then $kn=2r-1$ for some $r \in \mathbb{N}$ so $\sin{(k n \pi)} = \sin{(-\pi + 2\pi r)} = \sin{(-\pi)} = 0$ If $kn$ is even then $kn=2r$ for some $r \in \mathbb{N}$ so $\sin{(k n \pi)} = \sin{(0+2\pi r)} = \sin{(0)}=0$ Thanks from topsquark and idontknow November 1st, 2019, 11:41 AM #3 Senior Member   Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math What about this ? $\displaystyle s=\displaystyle \sum_{k=1}^{\infty } \frac{\cos(kn\pi )}{k}\; , n\in \mathbb{N}.$ November 1st, 2019, 12:17 PM #4 Senior Member   Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 Math Focus: Algebraic Geometry If $n$ is even, $n=2r$ for some $r \in \mathbb{N}$ so $\cos{(kn \pi)} = \cos{(2kr \pi)} = 1$ so the series becomes the harmonic series $\displaystyle \sum_{k=1}^{\infty}\frac{1}{k}$ which is divergent. If $n$ is odd, $\cos{(kn \pi)} = \cos{(k \pi (2r-1))} = \cos{(-k\pi + 2kr \pi)} = \cos{(-k \pi)} = \cos{(k \pi)}$ This turns the series into $\displaystyle \sum_{k=1}^{\infty} \frac{(-1)^k}{k}$ by the alternating series test this is convergent. According to Wolfram this series equals $-\ln{(2})$, but I don't know why. Thanks from topsquark and idontknow Last edited by skipjack; November 2nd, 2019 at 12:03 AM. November 1st, 2019, 12:33 PM   #5
Senior Member

Joined: Mar 2015
From: Universe 2.71828i3.14159

Posts: 169
Thanks: 65

Math Focus: Area of Circle
Quote:
 Originally Posted by Greens If $n$ is odd, $\cos{(kn \pi)} = \cos{(k \pi (2r-1))} = \cos{(-k\pi + 2kr \pi)} = \cos{(-k \pi)} = \cos{(k \pi)}$ This turns the series into $\displaystyle \sum_{k=1}^{\infty} \frac{(-1)^k}{k}$ by the alternating series test, this is convergent. According to Wolfram, this series equals $-\ln{(2})$, but I don't know why.
$$-1 < x \leqslant 1 \implies \ln (1+x) = \sum\limits_{k=1}^\infty \frac{(-1)^{k+1} x^k}{k}$$

And for $x=1$, $\displaystyle \ln(2) = \sum\limits_{k=1}^\infty \frac{(-1)^{k+1}}{k}=-\sum\limits_{k=1}^\infty \frac{(-1)^{k}}{k}$

Last edited by skipjack; November 2nd, 2019 at 12:15 AM. November 1st, 2019, 12:54 PM   #6
Senior Member

Joined: Mar 2015
From: Universe 2.71828i3.14159

Posts: 169
Thanks: 65

Math Focus: Area of Circle
Quote:
 Originally Posted by tahirimanov19 $$\ln (1+x) = \sum\limits_{k=1}^\infty \dfrac{(-1)^{k+1} x^k}{k}$$
Let's prove it.

$P(x)=\ln(1-x) \Rightarrow P'(x)= \dfrac{-1}{1-x} = -1 \cdot \left( 1+x+x^2+x^3+x^4+... \right)$

Therefore, $$P(x) = -x - \frac{x^2}{2}- \frac{x^3}{3}-... = - \sum\limits_{k=1}^{\infty} \frac{x^k}{k}$$

Plug $-x$ instead of $x$ and we get $$\ln (1- (-x)) = x - \frac{x^2}{2}+ \frac{x^3}{3}-... = \sum\limits_{k=1}^\infty \frac{(-1)^{k+1} x^k}{k}$$

So, $$\ln(1+x) = \sum\limits_{k=1}^\infty \frac{(-1)^{k+1} x^k}{k}$$

Last edited by skipjack; November 2nd, 2019 at 12:17 AM. November 1st, 2019, 01:06 PM #7 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle You can use just $$\ln (1-x) = - \sum\limits_{k=1}^{\infty} \frac{x^k}{k} .$$ For $x=-1$, $\displaystyle \ln(2)=- \sum\limits_{k=1}^{\infty} \frac{(-1)^k}{k}$. Do you have other infinite sum problems? Thanks from idontknow Last edited by skipjack; November 2nd, 2019 at 12:19 AM. November 1st, 2019, 11:27 PM #8 Senior Member   Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math $\displaystyle S=\sum_{k=1}^{\infty }e^{-k} 2^{k}$. November 2nd, 2019, 12:32 AM #9 Global Moderator   Joined: Dec 2006 Posts: 21,124 Thanks: 2332 $\displaystyle \sum_{k=1}^\infty e^{-k}2^k = \frac{1}{1 - 2/e} - 1 = \frac{2}{e - 2}$ Thanks from idontknow November 2nd, 2019, 12:49 AM   #10
Senior Member

Joined: Dec 2015
From: Earth

Posts: 833
Thanks: 113

Math Focus: Elementary Math
Quote:
 Originally Posted by skipjack $\displaystyle \sum_{k=1}^\infty e^{-k}2^k = \frac{1}{1 - 2/e} - 1 = \frac{2}{e - 2}$
Can you post the method in short-terms ? Tags compute, infinite, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post idontknow Real Analysis 4 October 21st, 2019 11:03 PM poopeyey2 Advanced Statistics 6 July 10th, 2017 01:12 PM poopeyey2 Advanced Statistics 7 July 3rd, 2017 11:42 AM Shamieh Calculus 9 November 16th, 2013 05:53 PM momo Number Theory 0 April 18th, 2009 11:09 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      