Question

I'm not quite sure what to do with this sum
\[\sum_{n=0}^{\infty} \frac{n!}{2*5*8*...*(3n+2)}\]

Answer 1

are you trying to evaluate that sum or finding the bound for it??

Answer 2

i'm supposed to determine if it's convergent or divergent

Answer 3

does this mean \[ \frac{n!}{\infty (3n+2)}\]

Answer 4

try using comparison test.

Answer 5

sorry my above statement is incorrect

Answer 6

the one with \[\infty\]

Answer 7

would my comparison test be
\[\sum_{n=0}^{\infty} \frac{n!}{(3n+2)}\]

Answer 8

if you remove that 0 at the lower limit you will get
\[ \frac{n!}{2*5*8*...*(3n+2)} \leq \frac{n!}{3*6*9*...*(3n)}\]

Answer 9

why?

Answer 10

\[\sum_{n=0}^{\infty} \frac{n!}{(3n+2)}\]
would this do me any good?

Answer 11

No wait
\[ \sum_{n=0}^{\infty} \frac{n!}{2.5.8...(3n+2)} = 1/2 + \sum_{n=1}^{\infty} \frac{n!}{5.8...(3n+2)} \leq 1/2 + \sum_{n=1}^{\infty} \frac{n!}{3.6.9...(3n+2)}\]

Answer 12

\[ \sum_{n=0}^{\infty} \frac{n!}{2.5.8...(3n+2)} = 1/2 + \sum_{n=1}^{\infty} \frac{n!}{5.8...(3n+2)} \leq 1/2 + \sum_{n=1}^{\infty} \frac{n!}{3.6.9...(3n)} \]

Answer 13

\[ \sum_{n=1}^{\infty} \frac{n!}{3.6.9...(3n+2)} = \sum \frac{n!}{3^nn!} = \sum \frac 1{3^n
}\]
which converges by comparison test.

Answer 14

how about the ratio test?
or would that be silly?

Answer 15

because that would eliminate the 2 5 8 etc

Answer 16

no not silly .. try as you like ... but this is simpler this way.

Answer 17

\[ \frac{(n+1)1}{2*5*8....etc}\ \frac{2*5*8....etc}{n!}\]

Answer 18

so it's convergent

Answer 19

\frac{(n+1)!}{2*5*8....etc}\ \frac{2*5*8....etc}{n!}
sorry typing error

Answer 20

\[\frac{(n+1)!}{2*5*8....etc}\ \frac{2*5*8....etc}{n!}\]

Answer 21

yeah ... you will get

looks like ratio test is simpler.