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If I take the power series around point $b$ of the following sum: $\sum_{i=2}^b \frac{\csc(i x)}{i}$

Series[Sum[Csc[i *x]/i, {i, 2, b}], {x, b, 3}] 

(mathematica)

I get this:

$\sum_{i=2}^b \left(\frac{\csc (b i)}{i}-(x-b) \cot (b i) \csc (b i)+(x-b)^2 \left(\frac{1}{2} i \csc (b i)+i \cot ^2(b i) \csc (b i)\right)+(x-b)^3 \left(-i^2 \cot ^3(b i) \csc (b i)-\frac{5}{6} i^2 \cot (b i) \csc (b i)\right)+O\left((x-b)^4\right)\right)$

This is useless as it has simply returned a more complicated version of the sum rather than a term by term approximation.

1) Why has this happened? and

2) What method (if any) can I use to get a term by term approximation like one gets for say sine about a point?

Appendum:

for sine about a point $\sum_{i=2}^b \sin(i x)$

Series[Sum[Sin[i *x], {i, 2, b}], {x, b, 1}] // TeXForm

$\frac{1}{2} \left(\cos \left(\frac{3 b}{2}\right) \csc \left(\frac{b}{2}\right)-\cos \left(b^2+\frac{b}{2}\right) \csc \left(\frac{b}{2}\right)\right)+\frac{1}{4} (x-b) \left(2 b \sin \left(b^2+\frac{b}{2}\right) \csc \left(\frac{b}{2}\right)+\sin \left(b^2+\frac{b}{2}\right) \csc \left(\frac{b}{2}\right)+\cos \left(b^2+\frac{b}{2}\right) \cot \left(\frac{b}{2}\right) \csc \left(\frac{b}{2}\right)-3 \sin \left(\frac{3 b}{2}\right) \csc \left(\frac{b}{2}\right)-\cos \left(\frac{3 b}{2}\right) \cot \left(\frac{b}{2}\right) \csc \left(\frac{b}{2}\right)\right)+O\left((x-b)^2\right)$

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    $\begingroup$ Have you tried computing the power series of the sine function about an arbitrary point? $\endgroup$ – saulspatz Jun 28 at 20:22
  • $\begingroup$ yes with $\sum_{i=2}^b \sin(i x)$ I get exactly what I want with Series[Sum[Sin[i *x], {i, 2, b}], {x, b, 3}] a series of terms and a Big Oh $\endgroup$ – onepound Jun 29 at 9:37
  • $\begingroup$ Are you talking about some computer program? $\endgroup$ – saulspatz Jun 29 at 12:47
  • $\begingroup$ @saulspatz the series was computed by mathematica but I'd imagine manually would produce the same result to your question 'Have you tried computing the power series of the sine function about an arbitrary point?'. $\endgroup$ – onepound Jun 29 at 13:04
  • $\begingroup$ I'm sorry, I completely missed your point. However, if you are talking about Mathematica, you should say that in the question. I know nothing of Mathematica, and wouldn't have ventured a comment. $\endgroup$ – saulspatz Jun 29 at 13:16

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