Series convergence - Gauss test How do I prove that
$$\sum_{n=1}^\infty\left(\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot(2n)}\right)^{k}$$
converges for $k&gt;2$ using Gauss test?
Easier: log the expression to get $k(\sum_{j=1}^{n} (2j+1) - \sum_{j=1}^{n} \log (2j))$, Here we have $n$ terms in both sums. After termwise subtraction we get $kn \log 1 = 0$. Hence the limit of the original series is $e^0=1$.
Using the notations of this presentation of the test, obviously, one is interested in the sequence
$$
u_n=\left(\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot(2n)}\right)^{k},$$
hence the job is to identify
$$
\frac{u_n}{u_{n+1}}=\ldots,
$$
then to identify $r\gt1$, $h$ and the bounded function $B$ such that
$$
\frac{u_n}{u_{n+1}}=1+\frac{h}n+\frac{B(n)}{n^r},
$$
and finally, to identify the values of $k$ such that $h\gt1$ (then the series converges) and the values of $k$ such that $h\leqslant1$ (then the series diverges).