I have an expression, and I wanted to evaluate it for a certain value of a variable. However, /. gives a calculation error, whereas manually replacing the variable with the value succeeds. Why?

$$\sum _{n=0}^{\infty } \frac{1}{(k+n)! (n-k)!}\text{/.}\, \{k\to 4\}$$ $$\sum _{n=0}^{\infty } \frac{1}{(n+4)! (n-4)!}$$

The first errors Infinity::indet: Indeterminate expression 0 ComplexInfinity encountered. and returns Indeterminate, whereas the second yields BesselI[8, 2]. My understanding of ReplaceAll was that the two should be functionally identical. What's wrong?

(Mathematica, Windows 10, 64 bit)

  • 2
    $\begingroup$ Hi Erhannis, welcome to the Mathematica Stack Exchange. Could you post the exact code that's exhibiting the issue in copy-pasteable format? It helps a lot if we can just plug your code straight into Mathematica. Thanks! $\endgroup$
    – Carl Lange
    Feb 2, 2019 at 8:09

1 Answer 1


The issue appears to be an order of precedence thing.

When you evaluate your first function $\sum _{n=0}^{\infty } \frac{1}{(k+n)! (n-k)!}$, it simplifies to $\frac{\, _1F_2(1;1-k,k+1;1)}{(-k)! k!}$.

Since it will do this simplification before replacing $k$, and since ${(-k)! k!}$ evaluates to ComplexInfinity (as does the top half of the function), you get the error message.

The trick is to do the replacement earlier:

$\sum _{n=0}^{\infty } \left(\frac{1}{(k+n)! (n-k)!}\text{/.}\, k\to 4\right)$

Which evaluates to BesselI[8, 2] as expected.

  • 3
    $\begingroup$ This kind of problem is exactly why HypergeometricPFQRegularized was introduced. Your general sum formula can be simplified to be non-singular: Sum[1/((n+k)!(n-k)!), {n,0,Infinity}] // FullSimplify gives HypergeometricPFQRegularized[{1},{1-k,1+k},1], which at k=4 gives Hypergeometric0F1Regularized[9,1], which in turn fullsimplifies to BesselI[8,2]. $\endgroup$
    – Roman
    Feb 2, 2019 at 8:56

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