Need hint regarding Euler's constant question

[librastar]

Homework Statement

Show that Euler's constant is 0 < \gamma < 1

Homework Equations

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

The Attempt at a Solution

At first glance I was thinking about proving by contradiction.

First I assume \gamma 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 \gamma does not equal 0.

Next I assume \gamma 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.

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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.

 

[gabbagabbahey]

Keep in mind that

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

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.

 

[Office_Shredder]

For proving the positivity of \gamma you should think about Riemann sums

 

[lanedance]

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

 

[Gib Z]

For proving the positivity of \gamma you should think about Riemann sums

Same for showing it is less than 1. Very simple and nice when you see the idea.