Absolute, conditional convergence or divergence

[kilroymcb]

I'm given: 1- ((1*3)/3!) + ((1*3*5)/5!) - ((1*3*5*7)/7!) ... + ((-1)^n-1) (1*3*5...(2n-1))/(2n-1)! +...
How do I even get started with this?
I suppose the + - alternation would indicate an alternating series. So, sum of (-1)^n (2n-1)/(2n-1)! and then do the ratio test?

 

[skeeter]

1 - ((1*3)/3!) + ((1*3*5)/5!) - ((1*3*5*7)/7!) + ... =

1 - 1/2 + 1/(2*4) - 1/(2*4*6) + ... + (-1)ⁿ/[2ⁿ*n!] + ...

any help?

 

[soroban]

Heello, kilroymcb!

Given:\(\displaystyle \L\:1\,-\,\frac{1\cdot3}{3!}\,+\,\frac{1\cdot3\cdot5}{5!}\,-\,\frac{1\cdot3\cdot5\cdot7}{7!}\,+ \cdots\,+\,\frac{(-1)^{n-1}\,\cdot\,1\cdot3\cdot5\,\cdots\,(2n-1)}{(2n\,-\,1)!}\)

First, we must construct the general term, \(\displaystyle a_n\).

How do we get: \(\displaystyle \:1\cdot3\cdot5\cdot7\,\cdots\,(2n-1)\,\) in the numerator?

Let's baby-talk our way through this . . .


Note that: \(\displaystyle \L\:1\cdot3\cdot5\cdot7\cdot9 \:=\:\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{2\cdot4\cdot6\cdot8}\)

. . \(\displaystyle \L=\:\frac{9!}{(2\cdot1)(2\cdot2)(2\cdot3)(2\cdot4)} \;=\;\frac{9!}{2^4(1\cdot2\cdot3\cdot4)} \;=\;\frac{9!}{2^4\cdot4!}\;\;\) This is the \(\displaystyle 5^{th}\) numerator

Hence, the general numerator is: \(\displaystyle \L\:\frac{(2n-1)!}{2^{n-1}\cdot(n-1)!}\)


Therefore, the general term is: \(\displaystyle \L\:a_n\;=\;\frac{\frac{(-1)^{n-1}(2n-1)!}{2^{n-1}(n-1)!}}{(2n-1)!} \;=\;\frac{(-1)^{n-1}}{2^{n-1}(n-1)!}\)

Now apply the Ratio Test . . .