Prove This Sequence Is Bounded Below by 0 Using Induction

[shephard23]

Homework Statement

Forgive my lack of LaTeX, not learned how to use it yet. Anyway, the problem is:

Use the inequalities

1/(n+1) < ln(n+1) - ln(n) < 1/n

to show that the sequence {xn} from n=1 to infinity defined by xn = 1 + 1/2 + ... + 1/n - ln(n) is strictly decreasing and bounded below by 0.

I've proved that it's strictly decreasing, stuck on the bounded below by 0 part.

Homework Equations

The Attempt at a Solution

We've been given a hint to use induction to show that xn > 0 for all natural numbers n so I'm going with that.

x(1) = 1 - ln(1) = 1 - 0 = 1 > 0. So it's true for n=1.

Assume it's true for n=k so x(k) = 1 + 1/2 + ... + 1/k - ln(k) > 0. I've tried rearranging this to 1 + 1/2 + ... + 1/k > ln(k) for use in the next step.

Now consider x(k+1) = 1 + 1/2 + ... + 1/k + 1/(k+1) - ln(k+1). I've to show this is greater than 0 if x(k) is greater than 0. The problem is that whenever I use the inequalities above I end up with x(k+1) is greater than something less than 0, for example:

1 + 1/2 + ... + 1/k + 1/(k+1) - ln(k+1)

> ln(k) + 1/(k+1) - ln(k+1) [using x(k)>0]

> 1(k+1) - 1/k [Using the inequality on the right above].

This is less than 0 so proves nothing, and most of my answers are coming out in this because I'm using similar methods. I thought I had to use the fact that x(k)>0 to show x(k+1)>0 so I've been trying it with no luck. Any advice at all?

 

[╔(σ_σ)╝]

This 1/(n+1) < log(n+1) - log(n) < 1/n inequality cannot possibly be correct .

Because from this you can deduce that
log(n+1) > log(n) + 1/(n+1)I tried log(100) - log (99) =0.004364

1/100 = .01

Already the inequality fails.

 

[shephard23]

My apologies, when we use log in our course it means the natural log. Sorry, I've been using it for so long it just became second nature. log in the above problem is the natural log, ln.

 

[shephard23]

I've managed to work out that x(k+1) > ln(k) - ln(k+1) + 1/k > 0 from the inequality on the right. Does anyone have any suggestions on getting

x(k+1) = 1 + 1/2 + ... + 1/k + 1/(k+1) > ln(k) - ln(k+1) + 1/k?

I'm having the same trouble as before.

 

[micromass]

You will need to show

log(n)<1+1/2+1/3+...+1/n

Let's use the inequality log(n+1)<log(n)+1/n.

We obtain

log(n)<log(n-1)+1/(n-1)<log(n-2)+1/(n-2)+1/(n-1)<...

I think you'll get there this way...