Calc question?

Question

If anyone could explain how the following problem is done, it would be greatly appreciated!

For the following series, use any suitable test to determine whether the series converges or diverges.

sigma starting at j=1 and going to infinity of:
(j^3 * 3^j) divided by j!

Answer 1

Use the ratio test...
limit of
[(j + 1)^3 * 3^(j+1) / (j+1)! ] / [j^3 * 3^j / j!] =

[ ( (j+1)^3 * 3^j * 3) * j! / (j+1)j! j^3 * 3^j ] =

[ ( (j+1)^3 * 3) / (j+1)j^3] =

[ ( (j+1)^2 * 3 / j^3)] =

and since this is <1, this sequence converges.

Answer 2

The series converges. You can compare succeeding terms to show this. If you have a term for the integer a and a+1, then the term for a is a^3 x 3^a/a!
the term for a+1 is
(a+1)^3 x 3^a x 3/ a!(a+1)
So the term(a+1)/term(a) = 3(a+1)^3 /(a+1)a^3
Then the ratio of terms is 3(a+1)^2/a^3.
which becomes 3.33 /a.
By comparison to the limiting divergent series SIGMA 1/j, term by term, this ratio is less than j/j+1, and so the series converges.