Hardy–Littlewood tauberian theorem

This formulation is from Titchmarsh.[1]^:226 Suppose a_n ≥ 0 for all n, and as x ↑1 we have

${\displaystyle \sum _{n=0}^{\infty }a_{n}x^{n}\sim {\frac {1}{1-x}}.}$

Then as n goes to ∞ we have

${\displaystyle \sum _{k=0}^{n}a_{k}\sim n.}$

The theorem is sometimes quoted in equivalent forms, where instead of requiring a_n ≥ 0, we require a_n = O(1), or we require a_n ≥ −K for some constant K.[2]^:155 The theorem is sometimes quoted in another equivalent formulation (through the change of variable x = 1/e^y ).[2]^:155 If, as y ↓ 0,

${\displaystyle \sum _{n=0}^{\infty }a_{n}e^{-ny}\sim {\frac {1}{y}}}$

then

${\displaystyle \sum _{k=0}^{n}a_{k}\sim n.}$

The following more general formulation is from Feller.[3]^:445 Consider a real-valued function F : [0,∞) → R of bounded variation.[4] The Laplace–Stieltjes transform of F is defined by the Stieltjes integral

${\displaystyle \omega (s)=\int _{0}^{\infty }e^{-st}\,dF(t).}$

The theorem relates the asymptotics of ω with those of F in the following way. If ρ is a non-negative real number, then the following statements are equivalent

• ${\displaystyle \omega (s)\sim Cs^{-\rho },\quad {\rm {{as\ }s\to 0}}}$
• ${\displaystyle F(t)\sim {\frac {C}{\Gamma (\rho +1)}}t^{\rho },\quad {\rm {{as\ }t\to \infty .}}}$

Here Γ denotes the Gamma function. One obtains the theorem for series as a special case by taking ρ = 1 and F(t) to be a piecewise constant function with value ${\displaystyle \textstyle {\sum _{k=0}^{n}a_{k}}}$ between t=n and t=n+1.

A slight improvement is possible. According to the definition of a slowly varying function, L(x) is slow varying at infinity iff

${\displaystyle {\frac {L(tx)}{L(x)}}\to 1,\quad x\to \infty }$

for every positive t. Let L be a function slowly varying at infinity and ρ a non-negative real number. Then the following statements are equivalent

• ${\displaystyle \omega (s)\sim s^{-\rho }L(s^{-1}),\quad {\rm {{as\ }s\to 0}}}$
• ${\displaystyle F(t)\sim {\frac {1}{\Gamma (\rho +1)}}t^{\rho }L(t),\quad {\rm {{as\ }t\to \infty .}}}$

The theorem can fail without the condition that the coefficients are non-negative. For example, the function

${\displaystyle {\frac {1}{(1+x)^{2}(1-x)}}=1-x+2x^{2}-2x^{3}+3x^{4}-3x^{5}+\cdots }$

is asymptotic to 1/4(1–x) as x tends to 1, but the partial sums of its coefficients are 1, 0, 2, 0, 3, 0, 4... and are not asymptotic to any linear function.

In 1911 Littlewood proved an extension of Tauber's converse of Abel's theorem. Littlewood showed the following: If a_n = O(1/n), and as x ↑ 1 we have

${\displaystyle \sum a_{n}x^{n}\to s,}$

then

${\displaystyle \sum a_{n}=s.}$

This came historically before the Hardy–Littlewood tauberian theorem, but can be proved as a simple application of it.[1]^:233–235

In 1915 Hardy and Littlewood developed a proof of the prime number theorem based on their tauberian theorem; they proved

${\displaystyle \sum _{n=2}^{\infty }\Lambda (n)e^{-ny}\sim {\frac {1}{y}},}$

where Λ is the von Mangoldt function, and then conclude

${\displaystyle \sum _{n\leq x}\Lambda (n)\sim x,}$

an equivalent form of the prime number theorem.[5]^:34–35[6]^:302–307 Littlewood developed a simpler proof, still based on this tauberian theorem, in 1971.[6]^:307–309