I don't think wolframalpha says it is divergent:

https://www.wolframalpha.com/input?i=sum%28binomial%28-2%2F5%2C+n%29%2C+%28n%2C+1%2C+inf%29%29

It does not compute the sum in closed form but gives a formula for the
partial sums as

  2^(-2/5)-1 + f(k)

where f(k) is something that goes to zero for large k.

SymPy gives the same partial sum formula in terms of 2F1 (hyper)
although it looks a little different with gamma functions:

In [29]: print(summation(binomial(S(-2)/5, n), (n, 1, k)))
(5/2 - 5*2**(3/5)/4)*gamma(3/5)/gamma(-2/5) - gamma(3/5)*hyper((1, k +
7/5), (k + 2,), -1)/(gamma(-k - 2/5)*gamma(k + 2))

I think SymPy and WolframAlpha are in agreement but just WA does not
compute a closed form for this particular sum whereas SymPy does get
the closed form but SymPy does not simplify the gamma functions as
nicely as WA does.

I think simplify here could be improved:

In [33]: e = summation(binomial(S(-2)/5, n), (n, 1, oo))

In [34]: print(e)
(5/2 - 5*2**(3/5)/4)*gamma(3/5)/gamma(-2/5)

In [35]: e.evalf()
Out[35]: -0.242141716744801

In [36]: 2**(-2/5)-1
Out[36]: -0.242141716744801

In [37]: print(simplify(e))
5*(2 - 2**(3/5))*gamma(3/5)/(4*gamma(-2/5))

Maybe gammasimp could handle this better somehow.

--
Oscar

On Wed, 10 Sept 2025 at 12:19, Paul Royik <[email protected]> wrote:
>
> sum of binomial(-2/5, n), n=1..infinity
>
> SymPy says that the answer is 2^(-2/5)-1.
> WolframAlpha says that the series is divergent.
>
> What answer is correct?
>
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