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 $kn=2r1$ 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 $\sin{(k n \pi)} = \sin{(0+2\pi r)} = \sin{(0)}=0$ 
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, $\cos{(kn \pi)} = \cos{(k \pi (2r1))} = \cos{(k\pi + 2kr \pi)} = \cos{(k \pi)} = \cos{(k \pi)}$ 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. 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:
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:
$P(x)=\ln(1x) \Rightarrow P'(x)= \dfrac{1}{1x} = 1 \cdot \left( 1+x+x^2+x^3+x^4+... \right)$ 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 $$\ln (1x) =  \sum\limits_{k=1}^{\infty} \frac{x^k}{k} .$$ For $x=1$, $\displaystyle \ln(2)= \sum\limits_{k=1}^{\infty} \frac{(1)^k}{k}$. Do you have other infinite sum problems? 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 
$\displaystyle \sum_{k=1}^\infty e^{k}2^k = \frac{1}{1  2/e}  1 = \frac{2}{e  2}$

November 2nd, 2019, 12:49 AM  #10 
Senior Member Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math  

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