Question

If $r<1$ and positive, and $m$ is a positive integer, show that $(2m+1)r^m(1-r)<1-r^{2m+1}$.
Hence show that $n{r}^n$ is indefinitely small when $n$ is indefinitely great.

Answer 1

The sum of a G.P. $l+r+r^2+r^3+\dots \dots +r^{2m}=\frac{1-r^{2m+1}}{1-r}$

Also, $(l-r^m{)}^2>0⟹1-2r^m+r^{2m}>0⟹1+r^{2m}>2r^m$

Similarly $r^p(l-r^{m-p}{)}^2>0⟹r^p-2r^m+r^{2m-p}>0⟹r^p+r^{2m-p}>2r^m$

Now, $l+r+r^2+r^3+\dots \cdots +r^{2m-2}{r}^{2m-1}+r^{2m}=(1+r^{2}m)+(r+r^{2m-1})+(r^2+r^{2m-2}+\dots \dots +r^m>(2m+1)r^m$

$\therefore (2m+1)r^m<\frac{1-r^{2m+1}}{1-r}⟹(2m+1)r^m(1-r)<1-r^{2m+1}$

Hence Proved

Multiplying both sides by $r^{m+1}$, we get

$(2m+1)r^{2m+1}(1-r)<r^{m+1}(1-r^{2m+1})$

Substituting $2m+1=n$, we get

$n{r}^n(1-r)<r^{\frac{n+1}{2}}(1-r^n)$

Making $n$ indefinitely great $r^{\frac{n+1}{2}}$ becomes indefinitely small, and therefore $n{r}^n$ becomes indefinitely small.