Just one more thing sir. Would you happen to know how much
the Sum of the Reciprocals of Odd Composites is?
See the footnotes for links to the prior related posts that lead up to this one.
I don’t know how this happened but recently I was thinking about the whole “Math Mystery” thing which ended up being all about coming up with the “formula” (value) for the
Sum Of The Reciprocals Of Composites which I represent as Sc.
That is, Sc, or the “Harmonic Series of Composites” (to coin a phrase)
is simply the log of the Prime Counting Function!
That is, Sc = ln(π(N)) or… Sc = ln( N / ln(N) )
Then it occurred to me that we now knew the following:
The Sum Of The Reciprocals Of All Numbers (aka the vanilla “Harmonic Series)
is equal to ln(N) which I will represent as Sn. So…
Sn = ln(N)
and The Sum Of The Reciprocals Of The Odd Numbers, as represented by So,
is 1/2 of the The Sum Of The Reciprocals Of The All Numbers. That is
So = Sn / 2
So = ln(N) / 2
And the Sum Of The Reciprocals Of Just The Primes is ln( ln(N) )
As I represent by Sp.
So, the Sum Of The Reciprocals Of The Odd Composites as represented by
Soc should be “All of the Odd reciprocals minus all of the Prime reciprocals
Soc = So – Sp
Soc = ln(N)/2 – ln(ln(N) )
But using some “log arithmetic” we can rewrite the above equation as
Soc = ln(N1/2) – ln( ln(N ) )
Using some more log arithmetic (See Footnote Log Arithmetic below) we get
Soc = ln(N1/2 / ln(N ) )
Or in English, the Sum Of The Reciprocals of just the Odd Composite numbers (aka Soc),
is the log of (the square root of N, divided by the log of N).
This is an interesting result when we consider that
the Sum Of The Reciprocals of ALL Composite numbers is
ln( N / ln(N) ).
SOddComposites = ln( N1/2 / ln(N)
SAllComposites = ln( N / ln(N) )
Who would have guessed.
Footnote on Log Rules:
1) logb(mn) = logb(m) + logb(n)
2) logb(m/n) = logb(m) – logb(n)
3) logb(mn) = n · logb(m)
Links to the first 4 episodes of the “Math Mystery”
which lead up to this post.