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I set the following to N=5, and want to do a convergence test on u:

    n = 5; u[r_, phi_, n_] := 
 Piecewise[{{BesselJ[1.5 r, n]*Exp[I n phi], 
    0 < r < 1/2}, {(BesselJ[3 r, n] + BesselY[3 r, n])*Exp[I n phi], 
    1/2 < r < 1}, {HankelH1[r, n]*Exp[I n phi], r > 1}}]

and the convergence test:

SumConvergence[u[r, phi, 5], n]

But the i get:

"SumConvergence::ivar: 5 is not a valid variable."

I try to set a number for it, but that is not accepted either.

How is this done correctly?

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  • $\begingroup$ In an infinite series $\sum_{k=1}^\infty a_k$, there is an index $k$ that ranges over (say, positive) integers. What is the index variable of your series? It's not 5 (therefore not n), since that is constant. The index should be the second argument of SumConvergence[]. $\endgroup$
    – Michael E2
    Commented Apr 21, 2021 at 12:29
  • $\begingroup$ That is a 0. I show here a successful use of this command: u0[r_, phi_] := Sum[I^(-n) BesselJ[n, r] Exp[I n phi], {n, 0, 3}]; u[r_, phi_] := u0[r, phi] SumConvergence[u0[r, phi], n] . Which is different from the one above. $\endgroup$
    – Superunknown
    Commented Apr 21, 2021 at 12:31
  • $\begingroup$ I get the ivar error with the code in your comment: i.sstatic.net/7ccuO.png -- Is that a successful use? (You didn't give an answer to "what is the index variable?". The constant 0 cannot be an index variable, but I'm not sure whether "That is a 0" was meant to be an answer.) $\endgroup$
    – Michael E2
    Commented Apr 21, 2021 at 12:43
  • $\begingroup$ This gives True: i.sstatic.net/ai4lk.png $\endgroup$
    – Michael E2
    Commented Apr 21, 2021 at 12:47
  • $\begingroup$ So, if you mean the first level of N, then it is 0. $\endgroup$
    – Superunknown
    Commented Apr 21, 2021 at 12:50

1 Answer 1

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In the general case SumConvergece fails as

Clear[n]; 
u[r_, phi_, n_] := Piecewise[{{BesselJ[3/2 r, n]*Exp[I n phi], 
0 < r < 1/2}, {(BesselJ[3 r, n] + BesselY[3 r, n])*Exp[I n phi], 
1/2 < r < 1}, {HankelH1[r, n]*Exp[I n phi], r > 1}}]
SumConvergence[u[r, phi, n], n,Assumptions -> r > 1/2 && phi \[Element] Reals]

demonstrates. It works if the parameters are specified, e.g

SumConvergence[u[2, Pi/4, n], n]

True

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