Convergence of $1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} - ...$ I was asked to prove whether
$$1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!}- \frac{1 \times 3 \times 5 \ \times 7}{7!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} - ...$$
converges absolutely, conditionally or diverges, and was wondering whether my proof is correct or not.
I began by noticing that
$$1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!}- \frac{1 \times 3 \times 5 \ \times 7}{7!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} -...$$
$$= 1- \frac{1}{2} + \frac{1}{2\times4} - \frac{1}{2 \times 4 \times 6}+...+\frac{1}{(2n)\times(2n-2)... \times2} -...$$
In other words, each $n$th term is the inverse of the product of all even natural numbers, and thus the initial sum is simply $ 1 + \sum a_n$ with
$$a_n := (-1)^n \frac{1}{\prod_{j=1}^n2j}$$
We can show that
$$\lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|} = \lim_{n\to\infty} \frac{\prod_{j=1}^{n}2j}{\prod_{j=1}^{n+1}2j}=\lim_{n\to\infty} \frac{1}{2(n+1)} = 0 < 1$$
and therefore the series converges absolutely according to the ratio test.
Is this proof correct?
 A: Even easier, we can explicitly evaluate the sum.  As you noticed, if for $n \ge 0$ $$a_n = \frac{(-1)^n}{(2n+1)!} \prod_{k=1}^n (2k+1),$$ then $$\begin{align}
a_n &= \frac{(-1)^n}{(2n+1)!} \prod_{k=1}^n (2k)(2k+1) \left(\prod_{k=1}^n (2k) \right)^{-1} \\
&= \frac{(-1)^n}{(2n+1)!} (2n+1)! \left( 2^n \prod_{k=1}^n k \right)^{-1} \\
&= \frac{(-1/2)^n}{n!}.
\end{align}$$
In fact, you did all of the above except the last step.
Therefore, $$\sum_{n=0}^\infty a_n = \sum_{n=0}^\infty \frac{(-1/2)^n}{n!} = e^{-1/2}.$$
If we just want to establish convergence, we can easily bound the sum by noting
$$|a_n| \le \frac{1}{n!}$$
for all $n$.
A: Possibly adding to the confusion, but
$$
x+{1\over 3!}x^3+{1\over 5!}x^5+...
$$
is the Taylor series for $\sinh x$ at $x=0$ and
$$
1+{1\over 2!}x^2+{1\over 4!}x^4+...
$$
is the Taylor series for $\cosh x$ at $x=0$ so your sum is
$$
\cosh\frac12-\sinh\frac12
$$
which is of course $\exp(-1/2)$ as noted in the other answer.
A: Yes, this is correct; there's nothing of note to mention otherwise that I can think of.

I'm choosing to answer like this and mark it as Community Wiki since I have nothing further to add, but don't want the question to end up in the unanswered queue.
