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Solve $$ \sum_{k = 1}^{ \infty} \frac{\sin 2k}{k}$$

I first tried to use Eulers formula

$$ \frac{1}{2i} \sum_{k = 1}^{ \infty} \frac{1}{k} \left( e^{2ik} - e^{-2ik} \right)$$

However to use the geometric formula here, I must subtract the $k=0$ term and that term is undefinted since $1/k$. I also end up with something that diverges in my calculations and since $\sin 2k$ is limited the serie should not diverge.

iveqy
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2 Answers2

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The summation is as follows: \begin{align} S &= \sum_{n=1}^{\infty} \frac{\sin(2an)}{n} = \frac{1}{2i} \, \sum_{n=1}^{\infty} \frac{ e^{2ai n} - e^{- 2ai n}}{n} \\ &= - \frac{1}{2i} \left( \ln(1 - e^{2ai}) - \ln(1 - e^{-2ai}) \right) \\ &= - \frac{1}{2i} \ln\left( \frac{1 - e^{2ai}}{1 - e^{- 2ai}} \right) = - \frac{1}{2i} \ln\left( - \frac{e^{ai}}{e^{-ai}} \cdot \frac{\sin(a)}{\sin(a)} \right) \\ &= - \frac{1}{2i} \ln\left( - e^{2ai} \right) = - \frac{1}{2i} \left( \ln(e^{- \pi i}) + \ln(e^{2ai}) \right) \\ &= - \frac{1}{2i} \left( - \pi i + 2ai \right) = \frac{ \pi - 2a}{2}. \end{align} This provides \begin{align} \sum_{n=1}^{\infty} \frac{\sin(2an)}{n} = \frac{ \pi - 2a}{2}. \end{align}

Leucippus
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  • Is there any reason to add an $a$ or is that just for clearity? And also, could you please explain how you did the step from $\sum$ to $\ln$? – iveqy Oct 05 '14 at 15:07
  • The sum S is pi-periodic with respect to a, not the formula at the end. Complex logarithms are many, one wonders which one to consider. A difference of logarithms is not always the logarithm of the ratio. And so on. – Did Oct 05 '14 at 15:07
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    For $a=2$, the final sum is off by $\pi$. There is definitely some care required with the logarithms. – Andrew Dudzik Oct 05 '14 at 15:30
  • @Slade ; Now, after some adjustment and use of Wolfram Alfa all is well. – Leucippus Oct 06 '14 at 04:27
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For finite non-zero $a,$

$$\sum_{k=1}^\infty\frac{e^{2iak}}k=-\ln(1-e^{2ia})$$

$$1-e^{2ia}=-e^{ia}(e^{ia}-e^{-ia})=-e^{ia}[2i\sin(a)]$$

$$\ln(1-e^{2ia})=\ln(e^{i(a+2m\pi)})+\ln2+\ln(-i)+\ln[\sin(a)]$$ $$=i(a+2m\pi)+\ln2-\frac{i\pi}2+\ln[\sin(a)]$$

as $-i=\cos\left(\frac\pi2\right)+i\sin\left(\frac\pi2\right)=e^{i\dfrac\pi2}$ and where $m$ is any integer

Hope you can take it home from here setting $a=1,-1$ one by one

lab bhattacharjee
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  • Where do you get that from? The Taylor expansion requires a $(-1)^{k-1}$ term. – iveqy Oct 05 '14 at 14:48
  • @iveqy, Please find the edited version, $$\ln(1-x)=-\sum_{r=1}^\infty\frac{x^r}r$$ – lab bhattacharjee Oct 05 '14 at 14:52
  • For $a=0$ this yields fascinating identities. Complex logarithms are many, one wonders which one to consider. That the logarithm of a product is the sum of the logarithms is certainly not true in general. And so on. – Did Oct 05 '14 at 15:04
  • @labbhattacharjee thankss, but I still don't understand why you add an $a$ and why how you go from $\sum$ to $\ln$. Could you elaborate on that? – iveqy Oct 05 '14 at 15:24
  • @iveqy, I have tried to generalize $1,-1$ with $a$. I've already commented $\ln(1-x)$ – lab bhattacharjee Oct 05 '14 at 15:27
  • So, after the patch of adding modulo $2i\pi$, which $m$ should we choose? Avoiding a proper treatment of the complex logs has its cost, in the end... – Did Oct 05 '14 at 20:39