Deduce if the series converges absolutely or conditionally.

Nov 2014
34
0
IceLand
[FONT=arial, sans-serif]Σ (-1)^k * 3^k(k!)^2/(2k)![/FONT]

[FONT=arial, sans-serif]I use ratio test.[/FONT]

[FONT=arial, sans-serif]|(3^k+1(k+1)!^2/(2(k+1)!/3^k(k!)^2/(2k)!|[/FONT]

[FONT=arial, sans-serif]which after simplifying got me [/FONT]

[FONT=arial, sans-serif]3(k+1)^2/(2k+2)(2k+1)[/FONT]

[FONT=arial, sans-serif]how does one show it converges absolutely?[/FONT]
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
You have \(\displaystyle \frac{3(k+ 1)^2}{(2k+ 2)(2k+ 1)}\). Multiplying out both numerator an denominator, that is
\(\displaystyle \frac{3k^2+ 6k+ 3}{4k^2+ 6k+ 2}\). Dividing both numerator and denominator by the highest power of k, \(\displaystyle k^2\),
we have \(\displaystyle \frac{3+ \frac{6}{k}+ \frac{3}{k^2}}{4+ \frac{6}{k}+ \frac{1}{k^2}}\).

What is the limit of that as k goes to infinity? What does that tell you? (You chose to use the ratio test, what does the ratio test say?)