Prove $a_n < a_{n + 1}$ if $b = 1 + \frac 1n$ and $a = 1 + \frac {1}{n + 1}$

Question

Let $b = 1 + \frac 1n$ and $a = 1 + \frac {1}{n + 1}$

Consider $\frac{b^{n + 1} - a^{n + 1}}{b - a} < (n + 1)b^n.$ we can substitute the given $a, b$ into this inequality to get

$\frac {(1 + \frac 1n)^{n + 1} - (1 + \frac {1}{n + 1})^{n + 1}}{b - a} < (n + 1)(1 + \frac 1n)^n$

which is

$\frac {a_nb -a_{n + 1}}{b - a} < (n + 1)a_n$

so that

$a_{n + 1} > (b -a)(n + 1)a_n - a_nb = a_n(-(bn - an - a))$

Does that work?

edit: I got a good answer on here, but I don't mind if this question is locked.

I am sorry I forgot to add that $a_n = (1 + \frac 1n)^n.$ – Jay Z Feb 22 '16 at 21:31 — Jay Z, Feb 22 '16 at 21:31

score 0 · Accepted Answer · answered Feb 23 '16 at 00:59

0

Yes! This does work.

It's usually preferred to state the full problem in the body of the post, even if it's in the title. Some people don't look at titles that much after viewing the post, and on the app it doesn't even render titles in MathJax

answered Feb 23 '16 at 00:59

Stella Biderman

31,155

Thank you very much. It really is a lot of help when someone confirms my work! Also, your suggestion is noted. – Jay Z Feb 23 '16 at 01:10

Prove $a_n < a_{n + 1}$ if $b = 1 + \frac 1n$ and $a = 1 + \frac {1}{n + 1}$

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