Von Mangoldt function

Question

Can anybody evaluate the following sum for me

$$ \sum\limits_{n=2}^\infty(-1)^n\left(\frac{\psi(n)}{n}-\frac{\Lambda(n)}{2n}\right) $$

where $\psi(n)$ is the Chebyshev function and $\Lambda(n)$ is the von Mangoldt function. Is there any efficient algorithm to evaluate such sum? I used the following code

f[n_]=Sum[MangoldtLambda[i],{i,1,n}];

Sum[(-1)^n(f[n]/n-MangoldtLambda[n]/(2n)),{n,2,Infinity}]

Answer 1

Number theory questions are always a huge accumulator for up votes. :)

From my experience I can say that the builtin MangoldtLambda function is pretty slow.

So let's define a Mangoldt function on our own. The Mangoldt function is defined by:

$\Lambda(n) \equiv \left\{ \begin{array}{1 1} \ln\ p & \quad \text{if n = $p^k$ for p a prime}\\ 0 & \quad \text{otherwise} \end{array} \right.$

This can be implemented as follows:

MangoldtΔ[n_] := If[Length[#] === 1, Log[#[[1, 1]]], 0] &[FactorInteger[n]]

Let's check some identities, if this implementation is correct:

$\sum_{d | n} \Lambda(d) = \ln\ n$

If we check this for n = 13:

Plus @@ (MangoldtΔ[#] & /@ Divisors[13])

==> Log[13]

Which is correct. Another identity is the following:

$\sum_{d | n} \mu(d)\ \ln(d) = -\Lambda(n)$ where $\mu$ is the Möbius function which is defined by:

$\mu(n) \equiv \left\{ \begin{array}{1 1} 0 & \quad \text{if $n$ has one or more repeated prime factors}\\ 1 & \quad \text{if $n=1$}\\ (-1)^k & \quad \text{if $n$ is a product of $k$ distinct primes} \end{array} \right.$

The first clause says nothing else but that n has to be square free:

MySquareFreeQ[n_] := Max[Last /@ FactorInteger[n]] < 2

(There is a SquareFreeQ function in Mathematica, so this is just if you're interested in how to implement such an algorithm)

Although there is a MobiusMu function in Mathematica we can define our own as well:

Moebiusμ[n_ /; MySquareFreeQ[n] == False] := 0
Moebiusμ[1] := 1
Moebiusμ[n_] := (-1)^Length@FactorInteger[n]

Now we can check this identity:

Plus @@ (Moebiusμ[n/#] Log[#] & /@ Divisors[13])

=> Log[13]

which is again correct.

Now let's do some timings with our new MangoldtΔ function:

Timing[Sum[MangoldtLambda[n], {n, 10^4}] // N]
=> {0.187940, 10013.4}

Timing[Sum[MangoldtΔ[n], {n, 10^4}] // N]
=> 0.065199, 10013.4}

We can see some improvement over time with our own implementation, but can we do better?

There is another identity with the Mangoldt which says:

$\psi(x) = \sum_{n \leq x} \Lambda(n)$

where $\psi(x)$ denotes the Chebyshev function. In my comments I already mentioned this function:

Chebyshevψ[x_] := Log[LCM @@ Range[x]]

Let's use the Chebyshevψ instead of Sum[Mangoldt...]:

Timing[ChebyshevPsi[10^4] // N]
=> {0.011974, 10013.4}

even better.

Equipped with these definitions and the proof of their correctness we can implement now your summation formula quite efficient:

Func[i_] := With[{iter = i}, 
    Sum[(-1)^n ((Chebyshevψ[n]/n) - (MangoldtΔ[n]/(2 n))), {n, 2, iter}]]


ListLinePlot[Table[Func[n], {n, 100}]]

Edit: Thank you Artes for pointing out the bug in (MangoldtΔ[n]/2 n)). Now this looks quite different.