Solve Sequence Problem: Prove 0<a<b Implies (n+1)bn > (b n+1 - a n+1)/(b-1)

[the_storm]

Homework Statement

Let a_n = ( 1 + \frac{1}{n} )ⁿ

Homework Equations

show that if 0 <= a < b

\frac{b ⁿ⁺¹ - a ⁿ⁺¹}{b-1} < (n+1)bⁿ

The Attempt at a Solution

I have started from a < b and I said so aⁿ < bⁿ

Then I multiply by (n+1) So I get the left hand side term.. but couldn't get the RHS .. any help guys?

 

[HallsofIvy]

Homework Statement

Let a_n = ( 1 + \frac{1}{n} )ⁿ

Homework Equations

show that if 0 <= a < b

\frac{b^{ n+1}- a^{n+1}}{b-1} &lt; (n+1)b^n

Do not use HTML "sup" or "sub" tags inside LaTeX. use "^" for superscripts, "_" for subscripts.

The Attempt at a Solution

I have started from a < b and I said so aⁿ < bⁿ

Then I multiply by (n+1) So I get the left hand side term.. but couldn't get the RHS .. any help guys?

I'm not sure what you are talking about. The inequality you are trying to prove does not seem to have any connection with the first limit you give. Is proving this a first step in proving that the limit exists? Also yoU say you started from a^n&lt; b^n then "multiply by (n+1)". How does that give you
\frac{b^{n+1}- a^{n+1}}{b- 1}

But my main problem is that inequality itself. I considered using induction but the inequality doesn't appear to be true for n= 0 or 1.

For example, for n= 0 it says that
\frac{b- a}{b- 1}&lt; 1
but that is not true for, say, b= 3, a= 1/2.

For n= 1,
\frac{b^2- a^2}{b- 1}&lt; 2b
which is not true for b= 3/2, a= 1/2.

 

[the_storm]

Ok I will use ^ and _ ... thank you.
concerning to the problem a and b must be integers only not fractions and the minimum value of the n is 1 because a_n is a sequence