I'm asking because it's a bit odd that I'm the only one posting. Should I hold off until the site is fixed? If that's the case, I'll stop with this question.
[math]\sum_{n=2}^{\infty}\frac{(n!)^2}{(2n)!}x^{2n+1}[/math]
I'm having some trouble solving this sum, particularly with the absolute value. Let me show you what I mean.
If I use the ratio test I get [imath]\displaystyle \lim_{n\to \infty}\left|\frac{((n + 1)!)^2}{(2(n + 1))!}x^{2(n + 1)+1}\frac{(2n)!}{(n!)^2x^{2n+1}}\right|[/imath]
I'll skip some simplification.
[math]\lim_{n\to \infty}\left|\frac{(n + 1)^2}{(2n+2)(2n+1)}x^2\right| = \left|\frac{1}{4}x^2\right|[/math]
The ratio test states that this converges only when [imath]\left|\frac{1}{4}x^2\right| < 1[/imath]
If I apply the absolute value rule along with the inequality, it leads to strange results.
[math]-1 < \frac{1}{4}x^2 < 1[/math]
I don't feel this inequality is correct because x^2 can't be negative.
[math]\sum_{n=2}^{\infty}\frac{(n!)^2}{(2n)!}x^{2n+1}[/math]
I'm having some trouble solving this sum, particularly with the absolute value. Let me show you what I mean.
If I use the ratio test I get [imath]\displaystyle \lim_{n\to \infty}\left|\frac{((n + 1)!)^2}{(2(n + 1))!}x^{2(n + 1)+1}\frac{(2n)!}{(n!)^2x^{2n+1}}\right|[/imath]
I'll skip some simplification.
[math]\lim_{n\to \infty}\left|\frac{(n + 1)^2}{(2n+2)(2n+1)}x^2\right| = \left|\frac{1}{4}x^2\right|[/math]
The ratio test states that this converges only when [imath]\left|\frac{1}{4}x^2\right| < 1[/imath]
If I apply the absolute value rule along with the inequality, it leads to strange results.
[math]-1 < \frac{1}{4}x^2 < 1[/math]
I don't feel this inequality is correct because x^2 can't be negative.