A question on summation notation and Raabe's test So I have a sum
$$\sum_{n=1}^\infty\frac{2\cdot4\cdot...\cdot2n}{5\cdot7\cdot...\cdot(2n+3)}$$
which I thought meant $$\sum_{n=1}^\infty\frac{2n}{2n+3}$$ which is trivially convergent by the limit test, and by Raabe's Test (which i'm currently practicing). But Taylor and Mann's book says this series is convergent. So I'm assuming These two series are not equivalent. If that is the case, how do I go about testing for convergence using Raabe's Test?
 A: The series
$$
\sum_{n=1}^\infty\frac{2n}{2n+3}\tag{1}
$$
diverges because the terms don't tend to $0$.

The ratio of the terms of the series
$$
\sum_{n=1}^\infty\frac{2\cdot4\cdot6\cdots2n}{5\cdot7\cdot9\cdots(2n+3)}\tag{2}
$$
is
$$
\frac{2n}{2n+3}\tag{3}
$$
Therefore,
$$
\begin{align}
\lim_{n\to\infty}\frac{2n}{2n+3}
&=\lim_{n\to\infty}\frac{2}{2+3/n}\\
&=1\tag{4}
\end{align}
$$
and
$$
\begin{align}
\lim_{n\to\infty}n\left(\frac{2n}{2n+3}-1\right)
&=\lim_{n\to\infty}\frac{-3n}{2n+3}\\
&=\lim_{n\to\infty}\frac{-3}{2+3/n}\\
&=-\frac32\\
&\lt-1\tag{5}
\end{align}
$$
$(4)$ and $(5)$, by Raabe's Test, indicate convergence.

As a check, we can also use the AM-GM Inequality to show
$$
\sqrt{(2n+2)(2n+4)}\le2n+3\tag{6}
$$
Therefore
$$
\begin{align}
\frac{2\cdot4\cdot6\cdots2n}{\color{#C00000}{5}\cdot\color{#00A000}{7}\cdot\color{#0000FF}{9}\cdots\color{#C08000}{(2n+3)}}
&\le\sqrt{\frac{2\cdot4\cdot6\cdots2n}{\color{#C00000}{4}\cdot\color{#00A000}{6}\cdot\color{#0000FF}{8}\cdots\color{#C08000}{(2n+2)}}}\sqrt{\frac{2\cdot4\cdot6\cdots2n}{\color{#C00000}{6}\cdot\color{#00A000}{8}\cdot\color{#0000FF}{10}\cdots\color{#C08000}{(2n+4)}}}\\
&=\sqrt{\frac{2}{2n+2}}\sqrt{\frac{2\cdot4}{(2n+2)(2n+4)}}\\
&\le\frac{\sqrt2}{(n+1)^{3/2}}\tag{7}
\end{align}
$$
then apply the $p$-test and comparison test to show that $(2)$ converges.
A: Hint: You have
$$\sum_{n = 1}^\infty a_n$$
where
$$a_n = \frac{2\cdot4\cdots2n}{5\cdot7\cdots(2n+3)}$$
To use Raabe's Test consider the quotient
$$b_n = \frac{a_{n + 1}}{a_n} = \frac{2\cdot4\cdots2(n+1)}{5\cdot7\cdots(2(n+1)+3)}\cdot\frac{5\cdot7\cdots(2n+3)}{2\cdot4\cdots2n} = \frac{2(n + 1)}{2(n + 1) + 3} = \frac{2n + 2}{2n + 5}$$
You have to compute the limit of $b_n$ and of $n(b_n - 1)$ as $n \to \infty$.
