Question

How would I find out if the following series converge or diverge absolutely or conditionally and find the sum if possible?

Answer 1

\[\sum_{n=1}^{\infty}n/(2n+3)\]
\[\sum_{n=1}^{\infty}n/(n^2 + 3)\]

Answer 2

n/2n +n/3
first one does not converge

Answer 3

Another one that I cannot figure out is \[\sum_{n=0}^{\infty}(n!)^2/(3n)!\]

Answer 4

When you have something like ! factorial, use ratio test

Answer 5

by ratio test, the limit seem to go to zero (l<1) , so converge
http://en.wikipedia.org/wiki/Ratio_test

Answer 6

I am a little confused about how to use the ratio test in this case, could you try and explain it for me.

Answer 7

Okay, here goes
\[\sum _{n=0}^{\infty } \frac{(n!)^2}{(3n)!}\]

limit of \[\frac{a_{n+1}}{a_n}\] is what we want
\[\frac{(n+1)!(n+1)}{3(n+1)!}*\text{ }\frac{(3n!)}{n! n!}\]
\[\frac{(n+1)^2}{(n+3)(n+2)(n+1)}\]
\[\frac{(n+1)}{(n+3)(n+2))}\]
Which goes to zero as n approach infinity

Answer 8

Now, how would I determine if it is absolute or conditional? I know it has something to do with L in relation to 1.

Answer 9

My understanding was you could only have conditional convergence when it is alternate series

Answer 10

Thank you very much for your help.

Answer 11

You are welcome