Does $a_n$ increasing imply $a_n-\frac{1}{n}$ strictly increasing?

Let $(a_n)_{n=1}^\infty$ be a sequence in $\mathbb{R}$. If $(a_n)$ is increasing prove that $(a_n-1/n)$ is strictly increasing.

How can I start off this off this question via induction?

• If $a_n=n+1/n$ then both $a_n$ and $a_n-\frac 1n$ are increasing. – lulu Mar 9 '16 at 17:25
• How is an=n+1/n ? – user321365 Mar 9 '16 at 17:32
• Also can someone tell me how to format. – user321365 Mar 9 '16 at 17:33
• To disprove a proposed theorem, it suffices to exhibit a single counterexample. I defined a particular sequence $a_n$ which, I believe, passes your requirements but which does not satisfy your conclusion hence the theorem, as stated, can not be correct. – lulu Mar 9 '16 at 17:36
• As to formatting, there is a terrific tutorial here: meta.math.stackexchange.com/questions/5020/… – lulu Mar 9 '16 at 17:37

$(\frac{1}{n})$ is strictly decreasing, so $(-\frac{1}{n})$ is strictly increasing. Then $a_{n+1}-\frac{1}{n+1} \geq a_{n}-\frac{1}{n+1} > a_{n}-\frac{1}{n}$.