Today I found a Sum which Mathematica will simplify for a general parameter value $\nu$, but which will not simplify fully for special cases $\nu = 1, 3, ...$ despite the fact that the general answer is correct for these special cases, as far as I can tell.

I was investigating an explicit form for $$ \frac{d}{d \nu} I_\nu(z) $$ Wolfram gives a series representation.

Plugging this into Mathematica I find:

In[23]:= FullSimplify[(Sum[(-((2^-\[Nu] z^\[Nu] Log[2])/
  Gamma[1 + k + \[Nu]]) + (2^-\[Nu] z^\[Nu] Log[z])/
  Gamma[1 + k + \[Nu]] - (
  2^-\[Nu] z^\[Nu] PolyGamma[0, 1 + k + \[Nu]])/
  Gamma[1 + k + \[Nu]]) (1/k! (z/2)^(2 k)), {k, 0, \[Infinity]}])]

Out[23]= BesselI[\[Nu], 
 z] (EulerGamma - HarmonicNumber[\[Nu]] + Log[z/2]) - 
 2^-\[Nu] z^\[Nu] Gamma[1 + \[Nu]] 
 RowBox[{"{", "1", "}"}], ",", 
 RowBox[{"0", ",", "0"}], "}"}], ",", "0"}], ")"}],
 MultilineFunction->None]\)[{1 + \[Nu]}, {1 + \[Nu], 1 + \[Nu]}, z^2/4]

But if I try the same thing for the special case $\nu =1$

Sum[(-((z Log[2])/(2 Gamma[2 + k])) + (z Log[z])/(2 Gamma[2 + k]) - (z PolyGamma[0, 2 + k])/(2 Gamma[2 + k])) (1/k! (z/2)^(2 k)), {k, 0, \[Infinity]}]]

It just returns the original sum. I have compared the original derivative of a Bessel function to the general expression for $\nu =1$ numerically and they seem to match. Why is it that Mathematica will not simplify these special cases?


The problem seems to be, that the sum does not converge for $\nu = 1$. I am not familiar with the mathematical background right now and do not have the time to do further investigations there, but here is what mathematica tells you:

Let's define the addend of your series separately:

term = ((-((2^-\[Nu] z^\[Nu] Log[2])/Gamma[1 + k + \[Nu]]) + (2^-\[Nu] z^\[Nu] Log[z])/Gamma[1 + k + \[Nu]] - (2^-\[Nu] z^\[Nu] PolyGamma[0, 1 + k + \[Nu]])/Gamma[1 + k + \[Nu]]) (1/k! (z/2)^(2 k)));

Then the following delivers the result you intended to obtain:

Sum[term, {k, 0, \[Infinity]}] /. \[Nu] -> 1 // FullSimplify


BesselI[1,z] (-1+EulerGamma+Log[z/2])-1/2 z (HypergeometricPFQRegularized^({1},{0,0},0))[{2},{2,2},z^2/4]

But mathematica assumed above that the series converges and then replaced $\nu$ with $1$. If you reverse the order mathematica tells you that the series does not converge:

Sum[term /. \[Nu] -> 1, {k, 0, \[Infinity]}] // FullSimplify

Sum::div: Sum does not converge.

| improve this answer | |
  • $\begingroup$ Odd that Mathematica does not give warnings or conditions for Sum the way it would with integrate. $\endgroup$ – Kevin Driscoll Oct 9 '13 at 16:54

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