Proving inequalities with logarithm

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  • #1
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I need to prove:

(n+1)*(log(n+1)-log(n) > 1 for all n > 0.

I have tried exponentiating it and I got

( (n+1)/n )^(n+1) < e.

And from there I couldn't go any farther, but I do know that it is true by just looking at its graph.

Could anybody help me please?
 

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  • #2
blue_leaf77
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Correction:
( (n+1)/n )^(n+1) > e.
As for the solution itself, first find out whether the function
$$
f(n) = \left( \frac{n+1}{n} \right)^{(n+1)}
$$
is a monotonically decreasing or increasing function for positive ##n## (it should be either of them). Then find the asymptote of this function as ##n \rightarrow \infty## in terms of ##e## by making use of the definition of ##e##
$$
e = \lim_{n \to \infty} \left( 1+\frac{1}{n} \right)^n .
$$
 
  • #3
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Correction:

As for the solution itself, first find out whether the function
$$
f(n) = \left( \frac{n+1}{n} \right)^{(n+1)}
$$
is a monotonically decreasing or increasing function for positive ##n## (it should be either of them). Then find the asymptote of this function as ##n \rightarrow \infty## in terms of ##e## by making use of the definition of ##e##
$$
e = \lim_{n \to \infty} \left( 1+\frac{1}{n} \right)^n .
$$

I tried that, but to know if it is indeed monotonic I have to show that the selected expression in the term the picture I attached is always < 1 or > 1.

So to proof of the original problem requires a proof for:
n*(log(n+1)-log(n)) < 1

Right now it looks pretty circular.
 

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  • #4
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I tried that, but to know if it is indeed monotonic I have to show that the selected expression in the term the picture I attached is always < 1 or > 1.
You don't need that, there is a more direct way.

You can also show the initial inequality directly with a clever integral. Can you rewrite ln(n+1)-ln(n) somehow?
 
  • #5
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You don't need that, there is a more direct way.

You can also show the initial inequality directly with a clever integral. Can you rewrite ln(n+1)-ln(n) somehow?
You don't need that, there is a more direct way.

You can also show the initial inequality directly with a clever integral. Can you rewrite ln(n+1)-ln(n) somehow?

Can I have more clues please? I'm getting nowhere plus I'm not that good with integral calculus since we were never taught this in my high school & university.
 
  • #6
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What is ##\int_a^b \frac{dx}{x}##?
 
  • #7
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What is ##\int_a^b \frac{dx}{x}##?
log(x), but I still really can't where I'm supposed to be headed.
 
  • #8
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It is not log(x). The definite integral not depend on x at all, but only on a and b.
 
  • #9
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It is not log(x). The definite integral not depend on x at all, but only on a and b.
log(b)-log(a) ?
 
  • #10
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Right, the difference between two logarithms - that's what you have in your original inequality.
 
  • #11
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Right, the difference between two logarithms - that's what you have in your original inequality.
I think you're implying that I integrate both sides, but how do I choose the value of a and b?
 
  • #12
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No, don't integrate. Replace (log(n+1)-log(n)) with an integral, and then find some way to get the inequality you need.
 
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