Can someone shed some light on this inequality?

Question

I have a question relating to image that I've attached. It is a proof that the sequence is increasing. I don't understand the logic behind the third equation $$\frac{a_{n+1}}{a_n}>\left (1-\frac{1}{n+1}\right ) \left (\frac{n+1}{n}\right)$$

where does the equation in the first and second parenthesis come from?

Ok, I have another relating question:

why $$\frac{a_{n+1}}{a_n}> (1+\frac{1}{n})$$ ( The expression of third line.

Answer 1

It is putting together the result from the first red box with the second one:

• $\frac{a_{n+1}}{a_n} = \color{blue}{\left(1- \frac{1}{(n+1)^2} \right)^{n+1}}\left( \frac{n+1}{n}\right)$
• $\color{blue}{\left(1- \frac{1}{(n+1)^2} \right)^{n+1}} > \color{green}{1 + (n+1)\left( \frac{-1}{(n+1)^2}\right)}$

$$\Rightarrow \frac{a_{n+1}}{a_n} > \left(\color{green}{1 + (n+1)\left( \frac{-1}{(n+1)^2}\right)}\right)\left( \frac{n+1}{n}\right) = \left(\underbrace{1- \frac{1}{n+1}}_{=\frac{n}{n+1}}\right)\left( \frac{n+1}{n}\right)$$

Answer 2

From Bernoulli's inequality, we have

$$\left( 1- \frac{1}{(n+1)^2}\right) > 1+(n+1) \left(\frac{-1}{(n+1)^2} \right)=1-\frac1{n+1}$$

Hence,

$$\frac{a_{n+1}}{a_n}>\left( 1- \frac{1}{(n+1)^2}\right)\left( \frac{n+1}{n}\right)>\left(1-\frac1{n+1} \right)\left( \frac{n+1}{n}\right)$$

Answer 3

So, we have $$\frac{a_{n+1}}{a_n} = \left(1 - \frac{1}{(n+1)^2}\right)^{n+1}\left(\frac{n+1}{n}\right).$$ The author then applies Bernoulli's inequality to the first term on the RHS: $$\left(1 - \frac{1}{(n+1)^2}\right)^{n+1} > 1 + (n+1)\left(\frac{-1}{(n+1)^2}\right) = 1 - \frac{1}{n+1}.$$ We can now return to the first equation and utilize this estimate; namely, we have $$\frac{a_{n+1}}{a_n} = \left(1 - \frac{1}{(n+1)^2}\right)^{n+1}\left(\frac{n+1}{n}\right) > \left(1-\frac{1}{n+1}\right)\left(\frac{n+1}{n}\right).$$ Finally, we multiply out the RHS of the inequality $$\frac{a_{n+1}}{a_n} > \left(1-\frac{1}{n+1}\right)\left(\frac{n+1}{n}\right) = \frac{n+1}{n} - \frac{1}{n} = 1.$$ So, we have $$\frac{a_{n+1}}{a_n} > 1 \implies a_{n+1} > a_n,$$ which means that $\{a_n\}$ is an increasing sequence.