# Euler's Constant Solution

1. May 12, 2003

### Tyger

In Mathematics a little substitution can work wonders. I'm going to show you how to solve a 150 year old riddle by using a little substitution.

Euler's Constant (Euler-Macheroni Constant), represented by &gamma; value .5772156649 has left mathematicians with two perplexing questions. Can it be represented as the sum of a series, and is it an irrational or transcendental number?

Normally &gamma; is represented as the limit of this expression:

(a) 1 + 1/2 + 1/3 ... +1/n &minus; ln(n)

however note that if we replace ln(n) with ln(n + 1/2)

the limiting value is approaced much more quickly. In fact any other substitution e.g. 2/5 or 3/5 doesn't approach the limit as rapidly. This turns out to have crucial significance for finding the answer for it allows us to change (a) so that each term in the harmonic series corresponds to a term in another series. This will result in &gamma; being represented as the difference between an infinite series and an infinite array, which answers our first question. Here is how each term appears:

(b) 1/n &minus; ln(n + 1/2) + ln(n &minus;1/2) which equals

(c) 1/n &minus; ln{(n &minus;1/2)/(n + 1/2)}

using the standard series for ln(x/y) which is

(d) 2&Sigma; {1/(2m + 1)}[(x &minus; y)/(x + y)]^2m + 1, for m = 0 and up.

Substituting for x and y and including the harmonic series term

(e) 1/n &minus; 2&Sigma;{1/(2m + 1)}(1/2n)^2m + 1

but note that the harmonic series term and the first term of the logarithmic series cancel. Also for n = 0 we have to add ln(2) so our final expression for the value of &gamma; is
&Sigma;
(f) ln(2) &minus; 2&Sigma;&Sigma; {1/(2m + 1)}(1/2n)^2m + 1 for m =1 and up and n = 1 and up.

Despite the fact that the array is two dimensional it converges rapidly. I haven't bothered to try to prove if it is irrational or transcendental, the former can be seen to be true almost by inspection, the latter should not be difficult to prove. I'm sure some young able-brained math whiz will have no trouble with it.

Last edited: May 12, 2003
2. May 13, 2003

### Hurkyl

Staff Emeritus
If you can see things like this by inspection, then you stand to become a very rich mathematician.

3. May 13, 2003

Staff Emeritus
What I want to know is, does that final formula converge to .5772156649 (at ten place accuracy). Anybody got some math software that will do the sums?

4. May 16, 2003

### Tyger

As I said

the array converges very rapidly so that a few minutes with a pocket calculator will get good accuracy. Also each row of the array is a logarithmic expression minus the harmonic term, so it can be summed row by row. It's also possible to appproximate the remainder of the array. These are all methods that were used before they had math programs.

In any case barring some clerical error on my part, it does give the correct value for Euler's Constant.

5. May 17, 2003

6. May 18, 2003