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Problem

Compute: $$S = \sum_{n=1}^{\infty} \frac{e^{-An}\sin(Bn)}{n}$$

My attempt, let $$ I = \sum_{n=1}^{\infty} \frac{e^{-An}}{n} \left( \cos(Bn) + i \sin(Bn) \right) = \sum_{n=1}^{\infty} \frac{e^{n(-A + iB)}}{n} $$

Using the series expansion,

$$ \sum_{x=1}^{\infty} \frac{e^{zx}}{x} = \log{\left( \frac{1}{1-e^{iz}} \right)} $$

we obtain

$$ I = \sum_{n=1}^{\infty} \frac{e^{n(-A + iB)}}{n} = \log{\left( \frac{1}{1-e^{i(-A + iB)}} \right)} $$

Which gives, $$ S = \Im(I) = \Im\left( \log{\left( \frac{1}{1-e^{i(-A + iB)}} \right)} \right). $$

However, I am not sure of this answer. This series originated while doing a problem of Physics, whose solution I do not possess. Any methods to do it more precisely are appreciated.

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1 Answer 1

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It seems that you have an extra $i$ $$I = \sum_{n=1}^{\infty} \frac{e^{n(-A + iB)}}{n} =-\log \left(1-e^{-A+i B}\right)$$

Expanding the complex numbers $$\Im(I)=\tan ^{-1}\left(\frac{\sin (B)}{e^A-\cos (B)}\right)$$ $$\Re(I)=\frac{1}{2} (A-\log (2 \cosh (A)-2 \cos (B)))$$

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  • $\begingroup$ Yes I accidentally put an $i$ n the Taylor expansion. $\endgroup$ Oct 1, 2023 at 12:21
  • $\begingroup$ How did you expand the complex numbers? $\endgroup$ Oct 1, 2023 at 12:21

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