# Why is $\sum_{k=0}^\infty \frac{k}{(1+r)^k} > 1/r$

Why is $$\sum_{k=0}^\infty \frac{k}{(1+r)^k} > 1/r$$ where $r$ is interest rate (small number)

Can someone give me a hint?

The Geometric series formula, for $-1<q<1$: $$\sum_{k=0}^{\infty}q^k=\frac{1}{1-q}$$ Derive by q $$\sum_{k=0}^{\infty}kq^{k-1}=\frac{1}{(1-q)^2}$$ Multiply by q $$\sum_{k=0}^{\infty}kq^{k}=\frac{q}{(1-q)^2}$$ Substitute $q=\frac{1}{1+r}$ $$\sum_{k=0}^{\infty}\frac{k}{(1+r)^{k}}=\frac{1+r}{r^2}=\frac{1}{r^2}+\frac{1}{r} >\frac{1}{r}$$

If $|x|<1$, then $$\sum_{k=0}^{\infty}kx^k=\frac{x}{(1-x)^2}$$