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 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
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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
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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
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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 ?

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