How to determine the $\sup (S) $ and $\inf (S) $ of $\frac {(n+1)^{2}}{ 2^{n} }$
the solution provided by the textbook is:
one can show by induction than for $n \ge 11, 2^{n} > (n+1)^{3}$. hence $ 0 < \frac{(n+1)^{2}}{(n+1)^{3}} = \frac {1}{n+1}$ for $n \ge 11$
therefore 0 is the greatest lower bound of our set.
it is also easy to show that$ 2^{n} < (n+1)^{2} $ for $n \ge 6$. hence $\frac {(n+1)^{2}}{2^{n}} < 1$ for $n \ge 6$. the numbers $2, \frac {9}{4}, \frac {25}{16}, \frac {36}{32}$ (greater than $1$) also belong to our set. thus the least upper bound of the set is $\frac {9}{4}$
i wish an explanation for the solution, explain why he did each step NOT how he did it
As this is not a do-it-for-me site, I'll only give a detailed hint for a starting point.
The simplest here is to determine whether $\;a_n=\dfrac{(n+1)^2}{2^{n}}$ is an (ultimately) monotonic sequence. As it is a sequence of positive numbers, we can try to determine whether $\frac{a_{n+1}}{a_n}>1\;\text{ or } <1$. So compute $$\frac{a_{n+1}}{a_n}=\frac{(n+2)^2}{2^{n+1}}\cdot\dfrac{2^{n}}{(n+1)^2}=\frac12\frac{(n+2)^2}{(n+1)^2},$$ so $\:a_n>a_{n+1}\:$ if and only if $$(n+2)^2<2(n+1)^2.$$ Can you proceed?
On the textbook solution:
He first shows, using the squeeze theorem, that the sequence tends to $0$ as $n\to\infty$. As it is a sequence of positive number, $0$ is a lower bound, and it is the greatest lower bound, because the limit can be approached at whatever precision you wish.
Then, he shows that $a_n$ (with my notation) is less than $1$ for $n\ge6$, whereas the first $5$ terms are greater than $1$, hence the lest upperbound is the greatest of the first $5$ terms.