I need assiatance to understand a problem in Inequality

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The forum discussion centers on proving the inequality (1+1/n)^n < (1+1/n+1)^(n+1) for any positive integer n. The proof involves analyzing the expression b^(n+1) - a^(n+1) and understanding how the inequality (n+1)a^n < (b^n + ab^(n-1) + ... + a^n) < b^n is derived from the assumption a < b. The key takeaway is that the repeated application of the inequality a < b allows for the establishment of bounds on the sums involved in the proof.

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sabyakgp
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Hello Friends,

I at a loss to understand the parts of the following proof:
For any positive ineteger n, prove that:

(1+1/n)^n < (1+1/n+1)^n+1
a, b positive real numbers such that a < b

Proof:

b^n+1 - a^n+1 = (b-a)(b^n+ab^n-1+...+a^n)
I could not understand the following part:
By a repeated use of a < b
(n+1)a^n < (b^n+ab^n-1+...+a^n) < b^n
.
.
.
How the inequality equation (n+1)a^n < (b^n+ab^n-1+...+a^n) < b^n
is derived from a < b?
I can understand how (n+1)a^n < (n+1) b^n, but how (n+1)b^n is greater than (b^n+ab^n-1+...+a^n) and how it's greater than a^n?
Could anyone please help me?

Best Regards,
Sabya
 
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an < akbn-k < bn for each k, where 0≤k≤n, since a < b. There are n+1 terms in the sum, so the final inequalities hold.
 

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