Proving inequalities with logarithm

[japplepie]

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?

 

[blue_leaf77]

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 $e$
$$
e = \lim_{n \to \infty} \left( 1+\frac{1}{n} \right)^n .
$$

 

[japplepie]

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.

 

[mfb]

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?

 

[japplepie]

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.

 

[mfb]

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

 

[japplepie]

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

log(x), but I still really can't where I'm supposed to be headed.

 

[mfb]

It is not log(x). The definite integral not depend on x at all, but only on a and b.

 

[japplepie]

It is not log(x). The definite integral not depend on x at all, but only on a and b.

log(b)-log(a) ?

 

[mfb]

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

 

[japplepie]

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?

 

[mfb]

No, don't integrate. Replace (log(n+1)-log(n)) with an integral, and then find some way to get the inequality you need.