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# Ian H

• ### A DIVERSION for your entertainment?

We know that the harmonic series diverges

H = 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + ... = inf

The sum of reciprocals of square roots, namely

S = 1/√(1) + 1/√(2) + 1/√(3) + 1/√(4) + 1/√(5) + 1/√(6) + 1/√(7) + 1/√(8)

must also diverge, (because the denominators are all smaller).

But suppose we divide the inf sum S into two parts, like this

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

and let E be the sum with even terms, E = 1/√(2) + 1/√(4) + 1/√(6) + 1/√(8)

We have arranged that O + E = S, term by term, but consider the series √(2)E

√(2)E = √(2)/√(2) + √(2)/√(4) + √(2)/√(6) + √(2)/√(8) + ...

√(2)E = 1/√(1) + 1/√(2) + 1/√(3) + 1/√(4) = S = E + O, so now we have

O = [√(2) – 1]E ~ 0.4142E

This result tells us that the odd sum is less than the even sum! But odd terms of the form 1/√(2n - 1) are larger than corresponding even terms 1/√(2n) which means that this is impossible. They both diverge of course, but which sum to n terms is larger? Jacob Bernoulli remarked on this apparent paradox.

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A quick calculation comparing 1000 terms resolves the issue.

Σ1/√(2n) ~ 43.6999

Σ1/√(2n - 1) ~ 44.2936

Although they both head off to infinity, comparable odd terms are always larger, so that series O is always slightly ahead at any intermediate stage.

The apparent result O ~ 0.4142E was just wrong.

But can you explain why?

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 Errors6 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?

Mathematics7 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)