idempotent

@Diarytells This holds for all |a| < 1

@Diarytells There's a nice generalization to this problem:

zeta(n) for even n can be evaluated as a rational multiple of a power of pi. Then terms can be recombined. But this loses the visual structure of the recurrence relation.

We can obtain that series by simply multiplying out the series for gamma and sin.

@Diarytells Here's a cleaner evaluation of the limit:

@Diarytells Using the Mellin transform of cos and the power series for gamma and cos:

@CPierre67 Or alternatively, look up the Poisson distribution for large n.

@CPierre67 Look up Laplace's Method.

Stirling's formula and its connection to gaussians. Check this ⬇️

@CPierre67 Laplace and central limit theorem.

@op_idempotent Reference?

@CPierre67 The error on one side of the peak is somewhat compensated by the error on the other even for small n.

@op_idempotent Wait … the Stirling formula seems to be a very good approximation even for small ‘n’ … how so?

@CPierre67 There is a rigorous proof, but it does not fit in the margin.

@op_idempotent Just visual?