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.

Spoiler alert

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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?