easy to state -- hard to prove

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[damyankorena]

The answer is ζ(5) using the fundamental theorem of engineering

[adrianamor]

Next level would be “easy to state - impossible to prove”

[wynautvideos]

Do we know variations? Like Sum(n^3sin(n))^-1 or Sum(n^2sin^2(n))^-1 ?

[koendos]

I find it cool, that the sum of sin(n)/n goes to (pi-1)/2

[emanuellandeholm]

from math import sin

def flint_hills_gen(N):
for n in range(1, N):
denom = n ** 3 * (sin(n)) ** 2
yield 1 / denom

FH =
print(sum(FH))

[rotoboravtov]

From empirical observation, good rational approximations of pi yield 1/sin(n) ~ n. This makes the asymptotic worst case as 1/n, but good rational approximations become exponentially rare with large n and the series converges since the rest of the terms scale worse than 1/n.

[ospreytalon]

Well at n=103993, the summand has an approximate value of 2.43x10^(-6). It seems like intuitively the n^3 term would dwarf the 1/sin^2(n) past a certain point

[franksaved]

I'm curious, what does "how rational pi is" mean?

[RayArias]

Σ (n=1 to inf) 1/n³ × csc² n. So yeah, obviously when n happens to be close to the period of the cosecant (any integer multiple of pi), sin goes to 0, so then csc is going to approach inf or -inf (depending upon which side of the period you pop up at), so yeah, that's going to happen. It's not how big your n integer is, it's how close it is to the period of csc. This would even happen without the exponent on the csc.

[MyOneFiftiethOfADollar]

surprised this has not been compared to a known divergent series. Who is Flint Hill?

[Coffeyhandle]

Another fun sum!!!


1/sin^2(k pi/n) from k = 1 to n-1

its (n^2-1)/3

[gustavocortico]

Well but the singularities become more and more point-like for large x.

[emulateiam]

what about sum(n^-x * sin^-y(n), 1, inf)

does anything higher automatically converge because the n^ term is so big?

[agnivsarkar]

Wolfram alpha says it diverges idk why

[hamzalebbar]

We can took the absolute value of the serie and notice that 0=< |sin²(n)| =< 1, with that, we have the following identity : [ sum of n goes to 1 to +infinity of (1/(n³sin²(n))) ] =< [ sum of n goes to 1 to +infinity of (1/(n³×1)) ] and since [ sum of n goes to 1 to +infinity of (1/(n³×1)) ] is a Riemann's serie with p=3>1, [ sum of n goes to 1 to +infinity of (1/(n³×1)) ] converges so [ sum of n goes to 1 to +infinity of (1/(n³sin²(n))) ] is absolutely convergent