Series: $ \sum_{n = 1}^{\infty} \left( \prod_{i = 1}^n (2i - 1) \middle/ \prod_{i = 1}^n 2i \right)^k $ In Section 10.16 of Apostol's Calculus (Problem 18), the reader is asked to prove that the series
$$ \sum_{n = 1}^{\infty} \left( \prod_{i = 1}^n (2i - 1) \middle/ \prod_{i = 1}^n 2i \right)^k $$
converges if $k > 2$ and diverges if $k \leq 2$ using Gauss's test. In the text, Gauss's test is given as follows. If there is an $N \geq 1$, an $s > 1$, and an $M > 0$ such that $a_{n + 1}/a_n = 1 - A/n + f(n)/n^s$ for all $n \geq N$, where $|f(n)| \leq M$ for all $n > 0$, then $\sum a_n$ converges if $A > 1$ and diverges if $A \leq 1$. Letting $a_n$ be the terms of the indicated sequence, we use the binomial series to obtain
$$\begin{align}
   \frac{a_{n + 1}}{a_n} ~ = ~ & \left( \frac{2n + 1}{2n + 2} \right)^k \\
                 ~ = ~ & \left( 1 - \frac{1}{2n + 2} \right)^k && \\
                 ~ = ~ & \sum_{i = 0}^{\infty} {k \choose i} \left(\frac{-1}{2n + 2} \right)^i \\
                 ~ = ~ & 1 - \frac{k}{2}\left(\frac{1}{n + 1}\right) + \sum_{i = 2}^{\infty}{k \choose i}\left(\frac{-1}{2n + 2} \right)^i \\
                 ~ = ~ & 1 - \frac{k}{2}\left(\frac{1}{n}\right) + \left( \sum_{i = 2}^{\infty}{k \choose i}\left(\frac{-1}{2n + 2} \right)^i + \frac{k}{2}\left(\frac{1}{n(n + 1)}\right)\right) \\
  ~ = ~ & 1 - \frac{k}{2}\left(\frac{1}{n}\right) + \left(\left\{ n^2 \sum_{i = 2}^{\infty}{k \choose i}\left(\frac{-1}{2n + 2} \right)^i + \frac{k}{2}\left(\frac{n}{n + 1}\right)\right\} \middle/ n^2 \right)
 \end{align}$$
This has the form required for Gauss's test, with $A = k/2$ and $s = 2$. It remains to determine whether the absolute value of the bracketed expression is bounded by some positive number. We see that it is, since the absolute value of $kn/2(n + 1)$ is bounded by $|k|/2$. In addition, the absolute value of $n^2$ times the above series indexed at $2$ is bounded by some positive constant $\phi$. Thus, the absolute value of the function parameter to Gauss's test is bounded above by $\phi + |k|/2$ (work omitted). So the series converges if $k/2 > 1$, $i$.$e$., if $k > 2$, and diverges if $k/2 \leq 1$, $i$.$e$., if $k \leq 2$.
Here is my question. I am trying to work through this text using only those results previously introduced, and Apostol hasn't yet introduced the generalized binomial series. My conclusion is that this problem was supposed to be solved without using the binomial series, or facts about its convergence and divergence. So how would one go about solving this problem without expanding $(1 - 1/(2n + 2))^k$ using the binomial series, assuming that $k$ is an arbitrary real number? My only thought is that I should decompose $k$ into its integer and fractional parts, then use the binomial theorem for integral values. But then it's difficult to apply Gauss's test with negative values of $k$.

By the generalized mean value theorem, for $0<x<1$ we have
$$
\ln(1-x) = - x - \frac1{(1-\xi)^2} \frac{x^2}2 \quad\text{and}\quad e^{-x} = 1 - x + e^{-\eta} \frac{x^2}2
$$
for some $0\le\xi,\eta\le x$. In particular, for $0<x\le\frac12$ we have
$$
-x-2x^2 \le \ln (1-x) \le -x \quad\text{and}\quad 1-x \le e^{-x} \le 1 - x + x^2.
$$
Therefore for $0<x\le\frac1{2k}$ (we can assume $k\ge2$ and use the comparison test for smaller $k$),
$$
(1-x)^k = \exp(k \ln (1-x)) \le \exp(-kx) \le 1 - kx + k^2x^2
$$
and
$$
(1-x)^k = \exp(k \ln (1-x)) \ge \exp(-kx-2kx^2) \ge 1-kx-2kx^2.
$$
We conclude that
$$
\bigg( 1 - \frac1{2n+1} \bigg)^k = 1 - \frac k{2n+1} + \frac{g(n)}{(2n+1)^2}
$$
where $|g(n)| \le k^2$. One can easily convert this into the exact form required by Gauss's test.
