Prove that for every natural number $n$, there exists a natural number $m$ such that $\frac{1}{m}-\frac{1}{m+1} < \frac{1}{n}$

Question

I am not sure how to proceed with my proof. I can easily explain in words why this is so (just by maximizing the values of the LHS and RHS then noting the left will always decrease faster than the RHS due to denominator getting smaller), but I do not think this is mathematically valid and doesn't necessarily prove anything.

This is how I did it.

$\frac{1}{m} – \frac{1}{m+1} < \frac{1}{n}$

$=\frac{1}{m(m+1)} < \frac{1}{n}$

Now, I am sadly stuck and have no way to proceed. I am thinking to expand this inequality like $\frac{1}{m(m+1)} < \text{something} < \frac{1}{n}$

or

$\text{something} < \frac{1}{m(m+1)} < \frac{1}{n}$, but I am not sure how to go about this or if this is even the right approach.

Answer 1

You have $\frac{1}{m(m+1)} < \frac{1}{n}$, which is equivalent to $m(m+1) > n$. Surely you can find a value of $m$ that will make this true.