<s>[INST] Hall and Knight- Higher Algebra Problem 23, Example V. b 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 $nr^n$ is indefinitely small when $n$ is indefinitely great.
I have tried by taking $\frac{1-r^{2m+1}}{1-r}$ as the sum of the series $1 + r + r^2 ... + r^{2m}$ and then we may try to prove that $(2m+1)r^m > 1 + r + r^2 ... + r^{2m}$.
I have proved the inequality but am unable to prove the limit. The solution book says something which I do not understand. The solution given is
$(2m+1)r^m(1-r)<1-r^{2m+1}$
(multiplying by $r^{m+1}$ on both sides)
$=>(2m+1)r^{2m+1}(1-r)<r^{m+1}(1-r^{2m+1})$
putting $n=2m+1$
$=>nr^n(1-r)<r^{\frac{n+1}{2}}(1-r^{n})$
this much I understand but then it says that if $n$ is made indefinitely great, then $r^{\frac{n+1}{2}}$ becomes indefinitely small and $nr^n$ becomes indefinitely small.
 [/INST] You have
$$\begin{aligned}
\frac{1 - r^{2m+1}}{1-r}&= 1+r+ \dots +r^{2m}\\
&= r^m(\frac{1}{r^m} + \frac{1}{r^{m-1}} + \dots + \frac{1}{r} + 1+ r +r^2 + \dots + r^{m-1}+r^m)\\
&= r^m[1 + (r^m+\frac{1}{r^m}) + (r^{m-1} + \frac{1}{r^{m-1}}) + \dots +(r + \frac{1}{r})]\\
&> r^m(1 + \underbrace{2 + \dots + 2}_{m \text{ times}}) = (2m + 1)r^m
\end{aligned}$$
because for $0<x<1$ you have $x + \frac{1}{x} >2$.
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