How to get from one inequality to another

Question

I am trying to add an extra step in between the inequalities (1) and (2) in a proof for the below lemma to try to understand how (2) was achieved. I have managed to understand how to get to the first sum in (2), but I can't seem to get a middle step in between the second sum in (1) and the second sum in (2). Could someone please explain how we get $$\sum^{\infty}_{n=N+1}r^n|c_n| \leq \frac{M}{N}\sum^{\infty}_{n=N+1}r^n .$$

Lemma: Presume that the Abel means $A_r = \sum^{\infty}_{n=1}r^nc_n$ of the series $\sum^{\infty}_{n=1}c_n$ are bounded as $r$ tends to $1$ (where $r<1)$. Then, if $c_n= O(1/n)$, we have that the partial sums $S_N=\sum^N_{n=1}c_n$ are bounded.

Proof: Let $r =1-\frac{1}{N}$ and pick $M$ s.t. $n|c_n|\leq M$. Now, we find an approximation for the difference $$S_N-A_r = \sum^N_{n=1}(c_n -r^nc_n) - \sum^{\infty}_{n=N+1}r^nc_n$$ using $$1-r^n = (1-r)(1+r+...+r^{n-1}) \leq n(1-r).$$ Our approximation is:

\begin{equation*} \begin{split} |S_N-A_r| &\leq \sum^N_{n=1}|c_n|(1-r^n)+ \sum^{\infty}_{n=N+1}r^n|c_n| \;\; (1)\\ &\leq \sum^N_{n=1}|c_n|n(1-r)+ \\ &\leq M\sum^N_{n=1}(1-r) + \frac{M}{N}\sum^{\infty}_{n=N+1}r^n \;\; (2)\\ \end{split} \end{equation*}

Answer 1

For all $n \geq N + 1$, we have :

\begin{equation*} \begin{split} |c_n| &= n \frac{|c_n|}{n} \\ &\leq n \frac{|c_n|}{N}\\ |c_n| &\leq \frac{M}{N}\\ \end{split} \end{equation*}

The first inequality holds because $n \geq N$; the second one holds by choice of $M$. From this it follows that: $$\sum^{\infty}_{n=N+1}r^n|c_n| \leq \frac{M}{N}\sum^{\infty}_{n=N+1}r^n$$