Inequality $\frac{1}{k^2} < \frac{1}{k(k-1)}$

Question

How to démonstrate :

$\frac{1}{k^2} < \frac{1}{k(k-1)}$. With k integer > 2 I just know that ln(1+1/k) < 1/k but i don’t thank that it can ne useful

Answer 1

It's $$\frac{1}{k(k-1)}-\frac{1}{k^2}>0$$ or $$\frac{1}{k^2(k-1)}>0$$ or $$k>1$$ or $$k\geq2$$

Answer 2

$k^2>k(k-1)>0$ for $k>2$. Larger denominator gives small fraction, which shows that

$\frac{1}{k^2}<\frac{1}{k(k-1)}$.

Alternatively, you can also find the difference

$\frac{1}{k^2}-\frac{1}{k(k-1)}=\frac{k-1}{k^2(k-1)}-\frac{k}{k^2(k-1)}=\frac{-1}{k^2(k-1)}$, which is negative. Hence

$\frac{1}{k^2}-\frac{1}{k(k-1)}<0$, which gives

$\frac{1}{k^2}<\frac{1}{k(k-1)}$.

Answer 3

You are thinking too hard.

For positive values $n < m \iff \frac 1n > \frac 1m$. And $k- 1 < k$ so $k(k-1) < k*k = k^2$.