I think I have the correct answer to the problem specifically, but I'm a little unsure of my usage of the little-o notation, or if this is even what Apostol was suggesting to do. (Apostol does cover little-o with regard to Taylor series, he hasn't mentioned big-O yet). The question in the book (Apostol Calculus Vol. I, Section 10.16 #18) is
Use Gauss' test to prove that the series $\sum_{n=1}^\infty \left(\frac{1 \cdot 3 \cdot 5\cdots(2n-1)}{2 \cdot 4 \cdot 6 \cdots(2n)} \right)^k$ converges if $k>2$, and diverges if $k\le2$.
Gauss' test is given in this form: If there is an $N \ge 1$, an $s>1$, and an $M\ge0$ such that
$\frac{a_{n+1}}{a_n}=1-\frac A n + \frac {f(n)}{n^s}\qquad \forall n\ge N$ where $|f(n)|\le M$ for all n, then $\sum a_n$ converges if $A>1$ and diverges if $A\le 1$.
My answer: $\frac {a_{n+1}}{a_n}=\left(\frac{2n+1}{2n+2}\right)^k=\left(1-\frac 1 {2(n+1)}\right)^k$ Let $x=\frac 1 {n+1}$, then $\frac{a_{n+1}}{a_n}=\left(1-\frac 1 2 x\right)^k=1-\frac k 2x+o(x)\quad \text{as }x\to0^+$ I think that this is almost the answer, since it shows that
$\frac{a_{n+1}}{a_n}=1-\frac k {2n}+o\left(\frac 1 n\right)\qquad \text{ as }n\to +\infty$ and so I see the correlation with Gauss' formula, and the distinction for $k\le2$ and $k>2$ for divergence and convergence. For some $N$, we have that the function on the right is, at most, some $\epsilon>0$ less than $\frac{a_{n+1}}{a_n}$ but I'm unsure of how to formally proceed.