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Need hint regarding Euler's constant question

  • Thread starter librastar
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1. The problem statement, all variables and given/known data

Show that Euler's constant is 0 < [tex]\gamma[/tex] < 1

2. Relevant equations

According to my book, [tex]\gamma[/tex] = lim((1+1/2+....+1/n) - log n) as n approaches infinity

3. The attempt at a solution

At first glance I was thinking about proving by contradiction.

First I assume [tex]\gamma[/tex] is equal to 0.

Then I will get lim(1+1/2+....+1/n) = lim (log n) as n approaches infinity.

However 1+1/2+....+1/n is a harmonic series which diverges, so the equality does not hold.

So [tex]\gamma[/tex] does not equal 0.

Next I assume [tex]\gamma[/tex] is less than 0.

Since log function never yields negative result, this implies that lim(1+1/2+....+1/n) < lim(log n) as n approaches infinity.

Again due to the divergent nature of harmonic series, the inequality does not hold.

These are all I have for now.


I feel I have logic error in the attempt to solve the problem, and I hope that someone will give me a hint on the correct solution.


Homework Helper
Gold Member
Keep in mind that

[tex]\lim_{x\to a} [f(x)-g(x)]=\lim_{x\to a}f(x)-\lim_{x\to a}g(x)[/tex]

Is only true if both the limits on the RHS exist.

I haven't tried the problem yet, but you might consider using the squeeze theorem.


Staff Emeritus
Science Advisor
Gold Member
For proving the positivity of [tex]\gamma[/tex] you should think about Riemann sums


Homework Helper
just to add to you first post, log(n) also diverges, so the harmonic argument doesn't quite hold

Gib Z

Homework Helper
For proving the positivity of [tex]\gamma[/tex] you should think about Riemann sums
Same for showing it is less than 1. Very simple and nice when you see the idea.

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