Series problem

[kilroymcb]

I really appreciate your help, guys.
Sum from 0 to infinityof (n!)/2*5*8....(3n+2)

The way I did this was to make (n!)/(3n-1)(3n+2) then do the ratio test. Absolute value
(n+1)!/(3(n+1)+1)(3(n+1)+2) all times (3n-1)(3n+2)/(n!).

Then:
(n+1)(3n+1)(3n+2)/(3(n+1)+1)(3(n+1)+2).
Factor out a n+1 from numerator and denominato.
Then:
(3+1)(1+2)/(3+4)(3+5) = 12/56 which is <1 and so the series is absolutely convergent?

 

[pka]

When all of the terms are non-negative, is there any reason to use absolute value?

 

[kilroymcb]

When all of the terms are non-negative, is there any reason to use absolute value?

Correct me if I'm wrong.. but yes. The first term in the denominator is 3n-1.

 

[soroban]

Hello, kilroymcb!

You simplified the ratio incorrectly . . .

\(\displaystyle \L\sum^{\infty}_{n=0}\frac{n!}{2\cdot5\cdot8\cdots(3n+2)}\)

\(\displaystyle \L\frac{a_{n+1}}{a_n} \;=\;\frac{(n+1)!}{2\cdot5\cdot8\cdots(3n+2)(3n+5)} \,\cdot\,\frac{2\cdot5\cdot8\cdots(3n+2)}{n!}\)

. . \(\displaystyle \L= \;\frac{(n+1)!}{n!}\,\cdot\,\frac{2\cdot5\cdot8\cdots(3n+2)}{2\cdot5\cdot8\cdots(3n+2)(3n+5)} \;=\;\frac{n\,+\,1}{3n\,+\,5}\)

Divide top and bottom by \(\displaystyle n:\;\;\L\frac{1\,+\,\frac{1}{n}}{3\,+\,\frac{5}{n}}\)

Take the limit: \(\displaystyle \L\:\lim_{n\to\infty}\left(\frac{1\,+\,\frac{1}{n}}{3\,+\,\frac{5}{n}}\right) \;=\;\frac{1\,+\,0}{3\,+\,0} \;=\;\frac{1}{3}\)