Sum[Binomial[k+n-1, n]*Binomial[k, n-k]/k, {k, 1, n}]


Binomial[1,-1+n] HypergeometricPFQ[{1,1-n,1+n}, {3/2-n/2,2-n/2}, -(1/4)]

Running the output formula gives:

Indeterminate expression 0 ComplexInfinity encountered.

From the binomial sum we see that for n = 1,2,3,... the values are 1, 5/2, 22/3, 105/4, 511/5,... .

Also note that Table[Binomial[1,n-1], {n,1,6}] = 1,1,0,0,0,...

(Version: 11.2.0 for Linux x86 (64-bit) (September 11, 2017))


Observe that only the terms with $k\ge\lfloor{(n+1)\over2}\rfloor$ contribute to the sum. Now try this:

expr=Sum[Binomial[k+n-1, n]*Binomial[k, n-k]/k, {k, Floor[(n+1)/2], n}]



It produces the output:


and a formula in terms of the hypergeometric ${}_3F_2$ function,

$\frac{\binom{\left\lfloor \frac{n+1}{2}\right\rfloor }{n-\left\lfloor \frac{n+1}{2}\right\rfloor } \binom{n+\left\lfloor \frac{n+1}{2}\right\rfloor -1}{n} \, _3F_2\left(1,\left\lfloor \frac{n+1}{2}\right\rfloor -n,n+\left\lfloor \frac{n+1}{2}\right\rfloor ;-\frac{n}{2}+\left\lfloor \frac{n+1}{2}\right\rfloor +\frac{1}{2},-\frac{n}{2}+\left\lfloor \frac{n+1}{2}\right\rfloor +1;-\frac{1}{4}\right)}{\left\lfloor \frac{n+1}{2}\right\rfloor }$

which evaluates correctly in producing the above table.


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