I have to prove the following relations: sum_(x=a)^(b-1)1/x>=logb−loga sum_(x=a+1)^(b)(1)/(x)<=log b-log a

Colton Gregory

Colton Gregory

Open question

2022-08-20

x = a b 1 1 x and x = a + 1 b 1 x
I have to prove the following relations:
x = a b 1 1 x log b log a
x = a + 1 b 1 x log b log a
I tried to use the relation that a b 1 x d x = log b log a and for the second relation I obtained that a + 1 b 1 x d x = log b log ( a + 1 ) < log b log a and I think that could be sufficient to prove the relation. Instead, for the first relation i have some problems, because using the same method I don't obtain a good result, indeed: a b 1 1 x d x = log ( b 1 ) log a that is not greater than log b log a. What do you think? How can I resolve it?

Answer & Explanation

Cristofer Briggs

Cristofer Briggs

Beginner2022-08-21Added 8 answers

First, welcome to MSE, and great opening post!
Here's my hints on how to analyze this problem:
Consider left hand and right hand sums of 1 x . For our first inequality we are using lefthand sums with width of 1, so then, by the shape of the graph 1 x we get an overestimation of the integral so that
x = a b 1 1 x a b 1 x d x
Do you see what to do for the second example?
As for your approach to the problem, you were close, realizing that a + 1 b 1 x d x a b 1 x d x is similar to what I'm doing - it just seems like you were one step away!

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?