3
$\begingroup$

Please help me find the summation of following series under limit.

$$ \lim_{n \to \infty} \sum_{x = 1}^{n}{1 \over x}\,\cos\left(\left[x - 1\right]{\pi \over 3}\right) $$

Thank you. :)

$\endgroup$
5
$\begingroup$

Hint: The $\cos$ terms form a pattern for each six terms: $1, 1/2, -1/2, -1, -1/2, 1/2,$ and back to $1$. That would at least let you split the series up into ones that are more manageable.

$\endgroup$
  • $\begingroup$ Well,it does look like it may help the case but still I am unable to use it. The difficulty is arising because the sum is made up of product of two different type of progressions! So, if you can kindly provide a solution then it would be of greater help. Thank you! :) $\endgroup$ – Abir Mukherjee Feb 10 '14 at 18:11
5
$\begingroup$

Your sum is the real part of $$ e^{-i\pi/3}\sum_{k=1}^\infty\frac1ke^{ik\pi/3} $$ which is the series for $$ \begin{align} -e^{-i\pi/3}\log\left(1-e^{i\pi/3}\right) &=-\left(\frac12-i\frac{\sqrt3}{2}\right)\left(-i\frac\pi3\right)\\ &=\frac\pi6\left(\sqrt3+i\right) \end{align} $$


Note on the log

Since $e^{i\pi/3}$ is on the unit circle at angle $\pi/3$, we have this diagram

$\hspace{3.8cm}$enter image description here

Thus, $\left|1-e^{i\pi/3}\right|=1$ and $\arg\left(1-e^{i\pi/3}\right)=-\pi/3$. Therefore, $$ \log\left(1-e^{i\pi/3}\right)=0-i\frac\pi3 $$

$\endgroup$
  • 1
    $\begingroup$ I think your omitted line are $$e^{i\phi}=\cos\phi+i\sin\phi, 1-e^{2yi}=1-\cos2y-i\sin2y=2\sin^2y-2i\sin y\cos y$$ $$=-2i\sin y(\cos y+i\sin y)\implies \ln(1-e^{2yi})=\ln2+\ln(-i)+\ln(\sin y)+iy$$ here $2y=\frac\pi3$ $\endgroup$ – lab bhattacharjee Feb 10 '14 at 18:39
  • 1
    $\begingroup$ @labbhattacharjee: actually, I just drew the points $1$ and $e^{i\pi/3}$ and noted that because of the angle of $\pi/3$, they formed an equilateral triangle with the origin. This tells that $|1-e^{i\pi/3}|=1$ and $\mathrm{arg}(1-e^{i\pi/3})=-i\pi/3$. This tells that $\log(1-e^{i\pi/3})=-i\pi/3$. $\endgroup$ – robjohn Feb 10 '14 at 19:04
  • $\begingroup$ Thank you very much!It is a perfect and beautiful solution. :) $\endgroup$ – Abir Mukherjee Feb 11 '14 at 6:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.