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

which I represent asSum Of The Reciprocals Of CompositesS_{c}.

**That is, Sc,**** or **the “** Harmonic Series of Composites**” (to coin a phrase)

is simply the

**!**

*log of the Prime Counting Function*That is, S

_{c}= ln(π(N)) or… S

_{c}= ln( N / ln(N) )

`Then it occurred to me that we now knew the following:`

The Sum Of The Reciprocals Of **All ****N****umbers **(aka the vanilla “Harmonic Series)

is equal to **ln(N)** which I will represent as **S**_{n}. So…

**S**_{n}** = ln(N)**

and * The Sum Of The Reciprocals Of The Odd Numbers*, as represented by

**S**

_{o}

**,**

is 1/2 of the

*. That is*

**The Sum Of The Reciprocals Of The All Numbers****S _{o} = S_{n} / 2**

Or

**S**

_{o}= ln(N) / 2And the * Sum Of The Reciprocals Of Just The Primes* is ln( ln(N) )

As I represent by

**S**

_{p}

**.**

So, the *Sum Of The Reciprocals Of The Odd Composites *** **as represented by

**S**_{oc} should be “All of the Odd reciprocals minus all of the Prime reciprocals

S_{oc} = S_{o} – S_{p}

S_{oc} = ln(N)/2 – ln(ln(N) )

But using some “log arithmetic” we can rewrite the above equation as

S_{oc} = ln(N^{1/2}) – ln( ln(N ) )

Using some more log arithmetic (See Footnote Log Arithmetic below) we get

S_{oc} = ln(N^{1/2} / 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) ).

S_{OddComposites} = ln( N^{1/2} / ln(N)

**S _{AllComposites} = ln( N / ln(N) )**

Who would have guessed.

Footnote on Log Rules:

1) log_{b}(*mn*) = log_{b}(*m*) + log_{b}(*n*)

2) log_{b}(^{m}/_{n}) = log_{b}(*m*) – log_{b}(*n*)

3) log_{b}(*m** ^{n}*) =

*n*· log

_{b}(

*m*)

Links to the first 4 episodes of the “Math Mystery”

which lead up to this post.

1 https://itbegsthequestion.com/ an-actual-math-mystery/

2 https://itbegsthequestion.com/ math-mystery-followup/

3 https://itbegsthequestion.com/ math-mystery-wrap-up/

4 https://itbegsthequestion.com/epilogue-to-the-composite-chronicle/