Is the sum of this infinite series zero?

Question

The series starts:

1 - 1/2 - 1/3 + 1/4 - 1/5 + 1/6 - 1/7 - 1/8 + 1/9 + 1/10 - 1/11 - 1/12 - 1/13 + 1/14 + ...

The sign in front of 1/n is negative if the prime factorization of n has an odd number of  factors (counted with multiplicity) and is positive otherwise.  For example 12 = 2 * 2 * 3, so the prime factorization of 12 has 3 factors, so it is -1/12 in the sum.

Please provide proof that the infinite sum is zero, or explain why it is not zero.

kb · Accepted Answer

The key point for me is that this series is formally equal to the infinite product

Π_(p prime) (1 - 1/p + 1/p^2 - 1/p^3 + ...)

= Π_(p prime) (1/(1 - (-1/p)), via infinite geometric series

= Π_(p prime) p/(p+1).

(This should be easy to prove by truncating the product at some large N, use the distributive property, and then show that the error term between the resulting 'truncated' series and your series goes to 0 as N→∞.)

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Then, to show that the series equals 0, it suffices to show that

1/Π_(p prime) p/(p+1) = Π_(p prime) (p+1)/p is diverges to infinity.

To this end, let P(x) = Π_(p ≤ x) (p+1)/p = Π_(p ≤ x) (1 + 1/p).

Then, ln P(x) = Σ_(p ≤ x) ln(1 + 1/p).

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We next show that the related series Σ_(p prime) (1/p - ln(1 + 1/p))

is convergent.

We start with (via power series for ln(1 + t) with t = 1/p)

-ln(1 + 1/p) = 1/p - 1/(2p^2) + 1/(3p^3) - 1/(4p^4) + ... .

So, 1/p - ln(1 + 1/p) = 1/(2p^2) - 1/(3p^3) + 1/(4p^4) - ...

Not only is this positive, but this is bounded above by 1/(2p^2),

from the right side being a convergent alternating series.

This allows us to apply the Comparison Test to the series

Σ_(p prime) (1/p - ln(1 + 1/p)):

Σ_(p prime) (1/p - ln(1 + 1/p))

< Σ_(p prime) 1/(2p^2)

< Σ(n = 1 to ∞) 1/(2n^2), which converges, being a multiple of a convergent p-series.

Thus, Σ_(p prime) (1/p - ln(1 + 1/p)) is convergent.

Moreover, Σ_(p ≤ x) (1/p - ln(1 + 1/p))

= Σ_(p prime) (1/p - ln(1 + 1/p)) - Σ_(p > x) (1/p - ln(1 + 1/p))

= Σ_(p prime) (1/p - ln(1 + 1/p)) + O( Σ(n > x) 1/(2n^2) ), as above

= Σ_(p prime) (1/p - ln(1 + 1/p)) + O(1/x), via Riemann sums.

= A + O(1/x), where A = Σ_(p prime) (1/p - ln(1 + 1/p)), which we know converges.

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Therefore,

Σ_(p≤x) ln(1 + 1/p)

= -Σ_(p≤x) (1/p - ln(1 + 1/p)) + Σ_(p≤x) 1/p

= -(A + O(1/x)) + Σ_(p≤x) 1/p.

However, Σ_(p≤x) 1/p = ln(ln x) + B + O(1/ln x) for some constant B.

[See Theorem 4.12 in Apostol's "Introduction to Analytic Number Theorem" for a proof.]

So, Σ_(p≤x) ln(1 + 1/p) = B - A + ln(ln x) + O(1/ln x).

Exponentiate both sides, noting that e^t = 1 + O(t):

exp{Σ_(p≤x) ln(1 + 1/p)} = e^(B-A) e^(ln(ln x)) (1 + O(1/ln x))

==> P(x) = C ln x + O(1), where C = e^(B-A).

Finally, this shows that lim(x→∞) P(x) = ∞, and so the series in question equals 1/Π_(p prime) p/(p+1) = lim(x→∞) 1/P(x) = 0, as required.

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Fun problem!

Is the sum of this infinite series zero?

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