Consider a series like this:

$$\sum_{n=1}^{\infty} (c_n r^{2-n/2}+d_n r^{-n/2})$$


I want, now, to keep only the terms for which the power of $r$ is negative, i.e., the non diverging terms for $r\rightarrow\infty$, and send to zero all the c[n] and d[n] which correspond to a positive power of $r$.

After this, I would like to keep all the terms of the general sum that go to zero more slowly than $r^{2}$, e.g.the terms with $r^{-1}$, $r^{-1/2}$, with their respective coefficients. Notice that this can come from different $n$, so it is not enough in general to truncate the series.

How can I do this using Series? I can't manage it.

  • $\begingroup$ Like Sum[If[2 - n/2 >= 0, 0, c[n] r^(2 - n/2)] + d[n] r^(-n/2), {n, Infinity}]? $\endgroup$ – Dr. belisarius Mar 5 '15 at 17:42
  • $\begingroup$ @belisarius This is not practical when I have many different expressions as exponents. In the question I used a toy sum, but in general I would need to impose many more conditions on that If $\endgroup$ – usumdelphini Mar 5 '15 at 17:47

You can use Piecewise to filter out any components you do not desire:

(* your exclusion conditions *)
conds={2-n/2<0, (* possibly more *) };


and then

Sum[filter[conds] c[n]r^(2-n/2) + filter[conds] d[n]r^(-n/2),{n,Infinity}]

This can easily be expanded for any conditions you might encounter.

| improve this answer | |
  • $\begingroup$ It is a little but uncomfortable because it does not seem to work on the whole series. I mean, unless I multiply filter[cond] individually in front of each term that can give problems, it does not work. $\endgroup$ – usumdelphini Mar 5 '15 at 21:24
  • $\begingroup$ If for example I have Sum[c[n] r^(1-n/2)+d[n] r^(n/2-1),{n,Infinity}] this will give zero $\endgroup$ – usumdelphini Mar 5 '15 at 21:26
  • $\begingroup$ @usumdelphini: Sure you will have to filter all your expressions within the sum. I updated the answer accordingly. Were it not for an infinite sum, this would be totally different story. $\endgroup$ – Jinxed Mar 5 '15 at 22:41

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