Define θ(x)=∑p≤xlogp. To prove Bertrand's postulate it is sufficient to prove that θ(2x)−θ(x)>0 for sufficiently large x and check smaller values of x by hand.
Remarks:
- One could calculate the asymptotic expansion of θ(x) and then verify that θ(2x)−θ(x)>0. However, I think this is more difficult because knowing the asymptotic expansion is already closely related to the prime number theorem.
- I use the terminology "derivative" of f(x) for expressions like f(x)−f(x2)
It turns out that it is easier to treat the Möbius transform ψ(x)=∑∞n=1θ(x1/n). From bounds on "derivatives" of ψ(x) Ramanujan deduces bounds on the "derivatives" of θ(x) . I call this the Tauber step.
To obtain bounds on "derivatives" of ψ(x), Ramanujan uses the Möbius transform log⌊x⌋!=∑∞n=1ψ(xn) and uses bounds on "derivatives" of log⌊x⌋!, and another Tauber step. Finally, bounds on log⌊x⌋! are easy because one has Stirling's approximation.
Hence, the structure of the proof is
Remarks:
To obtain bounds on "derivatives" of ψ(x), Ramanujan uses the Möbius transform log⌊x⌋!=∑∞n=1ψ(xn) and uses bounds on "derivatives" of log⌊x⌋!, and another Tauber step. Finally, bounds on log⌊x⌋! are easy because one has Stirling's approximation.
Hence, the structure of the proof is
bounds onbounds onbounds on(derivatives of)T→(derivatives of)T→(derivatives of)log⌊x⌋!ψ(x)θ(x)
Remarks:
- Here T→ means the Tauber step.
- What I call the Tauber step is a simple version of the Landau-Ingham Tauberian theorem, see notes from Michael Müger for details on this theorem.
- In the next post I will follow Ramanujan's manipulations to calculate bounds in a simpler situation.
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