Prove this result related to progressions I'm stuck with the following problem from Higher Algebra by Hall and Knight:

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 no clue about the second part of the problem, but for now I've spent the whole afternoon trying to work out the first part -- no success.
Here are my two attempts:
1) Consider the sum of the sequence
$S = 1 + r + r^2 + ... + r^{2m}$
$\Rightarrow S = \dfrac{1-r^{2m+1}}{1-r}$
Now I can claim that the sum of these terms is greater than any of them ($m$ and $r$ being positive), so $S > r^m$. This leaves me with:
$\dfrac{1-r^{2m+1}}{1-r} > r^m$ 
Which is incomplete.
2) This time we consider the sequence:
$S = 1 + 3r + 5r^2 + ... + (2m-1)r^{m-1} + (2m+1)r^m$
$\Rightarrow rS = r + 3r^2 + ... + (2m-3)r^{m-1} + (2m-1)r^{m} + (2m+1)r^{m+1}$
$\Rightarrow (1-r)S = 1 + (2r + 2r^2 + ... + 2r^m) - r(2m+1)r^m$
(after some simplification)
$\Rightarrow (1-r)^2S = 1 + r -3r^{m+1} -2mr^{m+1}$
Now, I can claim that the LHS is greater than zero, which gives:
$2mr^{m+1} < 1 + r - 3r^{m+1}$
$\Rightarrow (2m+1)r^{m+1} < 1 + r - 2r^{m+1}$
etc. However, this too is of different form. 
Please provide some help!

You have $0 < r < 1$, so you can divide by $r^m(1-r) > 0$ and get the equivalent inequality
$$\begin{align}2m+1 &< \frac{1}{r^m} \frac{1-r^{2m+1}}{1-r} = \frac{1}{r^m}\left(1 + r + \dotsb + r^{2m-1} + r^{2m}\right)\\
&= \left(\frac{1}{r^m} + \frac{1}{r^{m-1}} + \dotsb + \frac1r + 1 + r + \dotsb + r^{m-1} + r^m \right)
\end{align}$$
to prove.
In that form, we can easily deduce it from
$$\left(x + \frac1x \geqslant 2\right) \land \left(x+\frac1x = 2 \iff x = 1\right),$$
since we can write
$$\frac{1}{r^m} + \dotsb \frac1r + 1 + r + \dotsb r^m = 1 + \sum_{k=1}^m \left(r^k + \frac{1}{r^k}\right) > 1 + m\cdot 2.$$
