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November 1st, 2019, 10:15 AM   #1
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Compute infinite sum

Compute: $\displaystyle \sum_{k=1}^{\infty } \frac{\sin(kn\pi )}{k}\; , n\in \mathbb{N}.$
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November 1st, 2019, 11:06 AM   #2
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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 $\sin{(k n \pi)} = \sin{(0+2\pi r)} = \sin{(0)}=0$
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November 1st, 2019, 11:41 AM   #3
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What about this ? $\displaystyle s=\displaystyle \sum_{k=1}^{\infty } \frac{\cos(kn\pi )}{k}\; , n\in \mathbb{N}.$
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November 1st, 2019, 12:17 PM   #4
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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.
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Last edited by skipjack; November 2nd, 2019 at 12:03 AM.
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November 1st, 2019, 12:33 PM   #5
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Quote:
Originally Posted by Greens View Post
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 $x=1$, $\displaystyle \ln(2) = \sum\limits_{k=1}^\infty \frac{(-1)^{k+1}}{k}=-\sum\limits_{k=1}^\infty \frac{(-1)^{k}}{k}$
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Last edited by skipjack; November 2nd, 2019 at 12:15 AM.
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November 1st, 2019, 12:54 PM   #6
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Originally Posted by tahirimanov19 View Post
$$\ln (1+x) = \sum\limits_{k=1}^\infty \dfrac{(-1)^{k+1} x^k}{k}$$
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, $$\ln(1+x) = \sum\limits_{k=1}^\infty \frac{(-1)^{k+1} x^k}{k}$$
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Last edited by skipjack; November 2nd, 2019 at 12:17 AM.
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November 1st, 2019, 01:06 PM   #7
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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?
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Last edited by skipjack; November 2nd, 2019 at 12:19 AM.
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November 1st, 2019, 11:27 PM   #8
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$\displaystyle S=\sum_{k=1}^{\infty }e^{-k} 2^{k} $.
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November 2nd, 2019, 12:32 AM   #9
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$\displaystyle \sum_{k=1}^\infty e^{-k}2^k = \frac{1}{1 - 2/e} - 1 = \frac{2}{e - 2}$
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November 2nd, 2019, 12:49 AM   #10
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Originally Posted by skipjack View Post
$\displaystyle \sum_{k=1}^\infty e^{-k}2^k = \frac{1}{1 - 2/e} - 1 = \frac{2}{e - 2}$
Can you post the method in short-terms ?
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