Proof of double inequation After simplifying an equation I got this:
$$
n*(\frac{y}{x})^{n-1}<1+\frac{y}{x}+(\frac{y}{x})^{2}+...+(\frac{y}{x})^{n-1}<n
$$
While 
$$
n\geq2 
$$
$$
x>y\geq0
$$
While n in N and x,y in Q.
It seems obvious that the right side is bigger than the middle, but how do I write a proof for that? Which method do I use both for proving the left inequation and the right one?
 A: Since $x>y$, we know that $\frac{x}y < 1$ (divide both sides by $y$). After multiplying with $\left(\frac{x}y\right)^{n-1}$, it then becomes obvious that $\left(\frac{x}y\right)^n < \left(\frac{x}y\right)^{n-1}$ for an arbitrary natural $n$. An inductive argument using the previous results will even yield $\left(\frac{x}y\right)^n < 1$ (for all $n > 0$).
We can use this to prove the right inequality in your question:
$$1 + \left(\frac{x}y\right) + \left(\frac{x}y\right)^2 + \dots + \left(\frac{x}y\right)^{n-1} < 1+1+\dots+1 = n.$$
A similar argument will prove the left inequality.
I don't see a nice way to prove both inequalities at once. I think you're better off proving them seperately.
A: Let $n \in \mathbb{N}$ and $x,y \in \mathbb{Q}$ such that : $n\geq 2$ and $x>y\geq 0$
Then $\forall k \in \left\lbrace1,....,n-1\right\rbrace$ : $$\dfrac{y}{x}^{n-1} < \dfrac{y}{x}^k < 1$$
So $$\displaystyle \sum_{k=0}^{n-1} \dfrac{y}{x}^{n-1} < \sum_{k=0}^{n-1} \dfrac{y}{x}^k < \sum_{k=0}^{n-1} 1$$
Finally : 
$$\displaystyle n\dfrac{y}{x}^{n-1} <  \sum_{k=0}^{n-1} \dfrac{y}{x}^k < n$$
A: You want to show
$n*(\frac{y}{x})^{n-1}<1+\frac{y}{x}+(\frac{y}{x})^{2}+...+(\frac{y}{x})^{n-1}<n
$.
Let $\frac{y}{x} = r$,
so $0 < r < 1$.
Your statement becomes
$nr^{n-1}
\lt \sum_{k=0}^{n-1}r^k
\lt n
$.
Since
$0 < r < 1$,
$r^j < r^k$
when $j > k$
so,
for $n \ge 2$,
$\sum_{k=0}^{n-1}r^k
=r^{n-1}+\sum_{k=0}^{n-2}r^k
\gt r^{n-1}+\sum_{k=0}^{n-2}r^{n-1}
= n r^{n-1}
$.
The right side is even simpler.
Since $0 < r < 1$,
we have
$0 < r^k < 1$
for $k \ge 1$
so
$\sum_{k=0}^{n-1}r^k
\lt \sum_{k=0}^{n-1}1
=n
$.
