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