- #1

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(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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- Thread starter japplepie
- Start date

- #1

- 93

- 0

(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?

- #2

blue_leaf77

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As for the solution itself, first find out whether the function( (n+1)/n )^(n+1)>e.

$$

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.

- #4

mfb

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You don't need that, there is a more direct way.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 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

mfb

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What is ##\int_a^b \frac{dx}{x}##?

- #7

- #8

mfb

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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

- #10

mfb

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Right, the difference between two logarithms - that's what you have in your original inequality.

- #11

- #12

mfb

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