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}]? $\endgroup$ Commented Mar 5, 2015 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 $\endgroup$ Commented Mar 5, 2015 at 17:47

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Commented Mar 5, 2015 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 $\endgroup$ Commented Mar 5, 2015 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
    Commented Mar 5, 2015 at 22:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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