Taylor series of $\sqrt{1-x}$ This is problem 6.7.6 in Abbott's Understanding Analysis 2nd ed. The section is called The Weierstrass approximation theorem.

a) Let $c_{n} = \frac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots 2n}$ for $n\geq 1$. Show $c_{n} < \frac{2}{\sqrt{2n+1}}$.
b) Use a) to show that $\sum_{n=0}^{\infty}a_{n}$ converges (absolutely, in fact) where $a_{n}$ is the sequence of Taylor coefficients generated in Exercise 6.7.4. $\left[a_{n}=\frac{\prod_{k=1}^{n}(2k-3)}{\prod_{k=1}^{n}(2k)}=\frac{-1\cdot 3\cdot 5\cdots (2n-3)}{2\cdot 4\cdot 6\cdots 2n}\right]$
c) Carefully, explain how this verifies that equation (1) $\left[\sqrt{1-x}=\sum_{n=0}^{\infty}a_{n}x^{n}\right]$ holds for all $x\in [-1,1]$.

With help of this MSE answer I was able to answer a). For c) I think the point is that if $\sum_{n=0}^{\infty}a_{n}$ converges absolutely, then the power series converges too, because $\sqrt{1-x}=\sum_{n=0}^{\infty}a_{n}x^{n}=\sum_{n=0}^{\infty}a_{n}$ at $x=1$. And for $x=-1$ we could use the alternating series to prove that $\sum_{n=0}^{\infty}a_{n}(-1)^{n}$ converges. I have already proved in a previous exercise that it converges for $x=(-1,1)$.
My problem is with b). I suppose that Abbott wants us to use the Weierstrass M-test and maybe a comparison test. However, $\sum_{n=0}^{\infty}c_{n}$ diverges, so I don't know how to go from there. In other words, I feel I didn't get much information from a). What am I missing?
 A: Using the hint given by Gary in the comments of the question and the MSE linked in the question, I will answer the whole exercise.
a) First, let's start with a lemma.
For all $k \in \mathbb{N}$, $\frac{k}{k+1} < \frac{k+1}{k+2}$ holds.
Proof:
We can write $k^{2} +2k < k^{2} + 2k + 1$. This implies $k(k+2) < (k+1)^{2}$. Dividing both sides of the inequalities by $k+2$ and $k+1$ gives us the desired result.
Proof of the main result:
Consider
$
\begin{align}
c_{n}^{2} &= \frac{\prod^{n}_{k=1}(2k-1)^2}{\prod^{n}_{k=1}(2k)^2} \\
&< \frac{\prod^{n}_{k=1}(2k-1)(2k)}{\prod^{n}_{k=1}(2k)(2k+1)} \quad \text{by the lemma}\\
&= \frac{\prod^{n}_{k=1}(2k-1)}{\prod^{n}_{k=1}(2k+1)} = \frac{1}{2n+1}
\end{align}
$
So $c_{n} < \frac{1}{\sqrt{2n+1}} < \frac{2}{\sqrt{2n+1}}$, as desired.
b) We can write $a_{n} = -\frac{c_{n}}{2n-1}$, because $\frac{c_{n}}{2n-1} = \frac{\prod^{n}_{k=1}(2k-1)}{\prod^{n}_{k=1}(2k)}\frac{1}{2n-1} = \frac{\prod^{n}_{k=1}(2k-3)}{\prod^{n}_{k=1}(2k)} = -a_{n}$.
So for all $n$, we have $a_{n} < \frac{1}{(2n-1)\sqrt{2n+1}} < b_{n} := \frac{1}{\sqrt{2n(2n-1)^{2}}}$. Let's introduce a third sequence $c_{n} = \frac{1}{\sqrt{2}n^{3/2}}$. If we prove that $\sum_{n=0}^{\infty} b_{n}$ converges, then by the comparison test $\sum_{n=0}^{\infty} a_{n}$ converges too.
We prove that $\sum_{n=0}^{\infty} b_{n}$ converges using the limit comparison test:
$
\begin{align}
\lim_{n\rightarrow \infty} \left\lvert \frac{c_{n}}{b_{n}} \right\rvert 
&= \lim_{n\rightarrow \infty} \left\lvert \frac{\sqrt{2n(2n-1)^{2}}}{\sqrt{2}n^{3/2}}\right\rvert \\
&= \lim_{n\rightarrow \infty} \left\lvert \frac{\sqrt{2n(2n-1)^{2}/n^{3}}}{\sqrt{2}n^{3/2}/n^{3/2}}\right\rvert \\
&= \lim_{n\rightarrow \infty} \left\lvert \frac{\sqrt{2n(4n^{2} - 4n+1)/n^{3}}}{\sqrt{2}}\right\rvert \\
&= \lim_{n\rightarrow \infty} \left\lvert \frac{\sqrt{8n^{3}/n^{3} - 8n^{2}/n^{3}+2n/n^{3}}}{\sqrt{2}}\right\rvert \\
&= \lim_{n\rightarrow \infty} \left\lvert \frac{\sqrt{8 - 8/n+2/n^{2}}}{\sqrt{2}}\right\rvert \\
&= \left\lvert \frac{2\sqrt{2}}{\sqrt{2}}\right\rvert \\
&= 2
\end{align}
$
Since $\sum_{n=1}^{\infty} c_{n}$ converges as it is a p-series with $p>1$, then $\sum_{n=1}^{\infty} b_{n}$ converges, proving that $\sum_{n=0}^{\infty} a_{n}$ converges too.
c) Let's return to the Taylor series of $\sqrt{1-x}$.
Let $-1<x<1$. Then we have that $\left\lvert \sum_{n=0}^{\infty} a_{n} x^{n}\right\rvert \leq \sum_{n=0}^{\infty} \left\lvert a_{n} x^{n} \right\rvert = \sum_{n=0}^{\infty} \left\lvert a_{n} \right\rvert \left\lvert x^{n} \right\rvert = \sum_{n=0}^{\infty} \left\lvert a_{n} \right\rvert$.
Since $\sum_{n=0}^{\infty} \left\lvert a_{n} \right\rvert$ converges, as proved before, then the power series converges. Proving the desired result.
