I was hoping someone could view my workings for the following series, and decide whether or not I am correct, any tips would be great if I have made an error I have tried to use images for clarity, seen as my last post was....crap.I have to decide whether the folllowing series are absolutely convergent,conditionlly convergent, or divergent.
. . .\(\displaystyle \L \mbox{1) }\, \, \sum_{n=1}^{\infty}\, \frac{(-1)^n\, (2n\, + \, 3)}{n^2}\)
For (1) I get Absolutely it is divergent using the "Limit comparison test", where I choose bn = 1/n (the divergent harmonic series) and thus the limit to infinity of an/bn = 2 > 0 indicating an is also divergent, hence my answer.
Therefore, I get series (1) as conditionally convergent, as it satisfies the conditions of the "Alternating Series Test" (ie there are 3 conditions)
. . .\(\displaystyle \L \mbox{2) }\, \, \sum_{n=4}^{\infty}\, \frac{(-1)^{n+1}\, 2^n\, (n!)^2}{(2n)!}\)
For (2) I get the Series is Absolutely convergent, my workings are displayed in teh following where I used the "ratio test":
. . .\(\displaystyle \L \frac{a_{n+1}}{a_n}\, =\, \frac{2^{n+1}\, \left(\, (n\, +\, 1)!\, \right)^2}{(2n\, +\, 2)!}\, \times\, \frac{(2n)!}{2^n\, (n!)^2}\)
. . . . . . . .\(\displaystyle \L =\, \frac{2\, (n\, +\, 1)\, (n\, +\, 1)}{2\, (n\, +\, 1)\, (2n\, +\, 1)}\)
. . . . . . . .\(\displaystyle \L =\, \frac{n\, +\, 1}{2n\, +\, 1}\)
Then:
. . .\(\displaystyle \L \lim_{n \rightarrow \infty}\,\, \frac{a_{n+1}}{a_n}\, =\, \frac{1}{2}\, \lt\, 1\)
Note: in both cases, I simply removed the (-1)^n component for absolute considerations.
. . .\(\displaystyle \L \mbox{1) }\, \, \sum_{n=1}^{\infty}\, \frac{(-1)^n\, (2n\, + \, 3)}{n^2}\)
For (1) I get Absolutely it is divergent using the "Limit comparison test", where I choose bn = 1/n (the divergent harmonic series) and thus the limit to infinity of an/bn = 2 > 0 indicating an is also divergent, hence my answer.
Therefore, I get series (1) as conditionally convergent, as it satisfies the conditions of the "Alternating Series Test" (ie there are 3 conditions)
. . .\(\displaystyle \L \mbox{2) }\, \, \sum_{n=4}^{\infty}\, \frac{(-1)^{n+1}\, 2^n\, (n!)^2}{(2n)!}\)
For (2) I get the Series is Absolutely convergent, my workings are displayed in teh following where I used the "ratio test":
. . .\(\displaystyle \L \frac{a_{n+1}}{a_n}\, =\, \frac{2^{n+1}\, \left(\, (n\, +\, 1)!\, \right)^2}{(2n\, +\, 2)!}\, \times\, \frac{(2n)!}{2^n\, (n!)^2}\)
. . . . . . . .\(\displaystyle \L =\, \frac{2\, (n\, +\, 1)\, (n\, +\, 1)}{2\, (n\, +\, 1)\, (2n\, +\, 1)}\)
. . . . . . . .\(\displaystyle \L =\, \frac{n\, +\, 1}{2n\, +\, 1}\)
Then:
. . .\(\displaystyle \L \lim_{n \rightarrow \infty}\,\, \frac{a_{n+1}}{a_n}\, =\, \frac{1}{2}\, \lt\, 1\)
Note: in both cases, I simply removed the (-1)^n component for absolute considerations.