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# Ian H

Consider the infinite series S

S = 1/√(1) + 1/√(2) + 1/√(3) + 1/√(4) + 1/√(5) + 1/√(6) + 1/√(7) + ......

Jacob Bernoulli knew that this sum of reciprocals of square roots, must

diverge since the denominators were all smaller than the harmonic series.

Let the inf sum of odd terms, O = 1/√(1) + 1/√(3) + 1/√(5) + 1/√(7) + .....

and the inf sum of even terms, E = 1/√(2) + 1/√(4) + 1/√(6) + 1/√(8) + ...

By inspection O + E = S, but we may also write √(2)E = S, so we have

O + E = √(2)E, or,

O = [√(2) – 1]E  ~ 0.4142E

Bernoulli remarked on the apparent paradox that the odd sum seems less than the even sum, but this is impossible because term by term odd terms are larger

Can you resolve the paradox with a clear explanation?

Notices and Errors4 months ago
• ### What s the sum from 1 to infinity of the product cos(pi/n)*cos(2pi/n)*cos(3pi/n)*...*cos(npi/n) .... How do we calculate this?

Mathematics5 months ago

• ### Complex integral around unit circle at origin for 1/z is 2𝛑i, For 1/z^n it is zero as per Cauchy s first even though pole there. Why?

The function f(z) = 1/z, is

not analytic at the origin; it has a type of singularity there called a pole.

Accepting that result and the reason given, it seemed reasonable that

if f(z) = 1/z^n, around a unit circle, then ∮(1/z^n)dz would also NOT be zero.

But the result of calculation is that ∮(1/z^n)dz = 0, (unless n = 1)