Calculate if an expression is uniformly lower than another?

Question

I have two expressions $$ \frac{2}{n-1} \quad \text{and} \quad \frac{2n-1}{n^2} $$ and I need to prove that $$ \frac{2n-1}{n^2} < \frac{2}{n-1} $$ for all $n \in \{2,3,\ldots\}$.

How can I do this?

Answer 1

Note that $n \ge 2$ to begin with. That said, then you want $\dfrac{2n-1}{n^2}- \dfrac{2}{n-1} < 0$. Rewrite this as a single fraction and you will see some thing new !

Answer 2

Multiply away the denominators, the way you would of this were an equation and not an inequality. Note that the denominators are always positive, so this doesn't affect the inequality sign.

Take the resulting inequality and tidy it up a bit (you will, for instance, have $2n^2$ on both sides that cancel one another). The inequality should now be a bit more manageable.

Answer 3

You need to clear the denominators carefully. By that I mean use the following property: $$ \frac{A}{B} < \frac{C}{D} \quad \Longleftrightarrow \quad AD < BC \quad \text{provided that } C>0 \text{ and } D>0 $$ Since $n \geq 2$, your denominators are positive: $n^2 > 0$ and $n-1 > 0$. So the inequality is equivalent to $$ (2n-1)(n-1) < 2n^2. $$ After expanding, you get $$ 2n^2 - 3n + 1 < 2n^2 $$ or $$ -3n + 1 < 0. $$ Finally, after rearranging a bit, $$ n > \frac{1}{3}, $$ which is certainly true as $n \geq 2$.