Does the series converge or diverge

sum m=1 to infinity (3^M)*(M^3) / M!

what test do i use?

A mixture of exponentials and factorials suggests the use of the Ratio Test, so use it.

With a_M = (3^M * M^3)/M!, the Ratio Test yields:
lim (M-->infinity) (a_M+1)/(a_M)
= lim (M-->infinity) {[3^(M + 1) * (M + 1)^3]/(M + 1)!}/[(3^M * M^3)/M!]
= lim (M-->infinity) [M! * 3^(M + 1) * (M + 1)^3]/[(M + 1)! * 3^M * M^3]
= lim (M-->infinity) [M!/(M + 1)! * 3^(M + 1)/3^M + (M + 1)^3/M^3]
= 3 * lim (M-->infinity) 1/(M + 1) * lim (M-->infinity) (M + 1)^3/M^3
= (3)(0)(1)
= 0.

Since |lim (M-->infinity) (a_M+1)/(a_M)| = 0 < 1, the series converges.

I hope this helps!

The above poster is right: just want to give you some intuition on this.

M^3 diverges much, much slower than 3^M. In fact, it is so much slower that we can pretty much pretend it's not there. [This would be okay if we think it might converge, also: since if the bottom converges with fast on top, it usually converges with fast+slow on top. And of course, if it diverges with fast on top, then surely it diverges with fast+slow on top!]

It's harder to see why, but 3^M diverges quite a bit slower than M!. So the whole thing converges.

Then, yeah, you use the ratio test to confirm that hypothesis.

when you have exponents, factorials, powers of the variable usually you want to use the ratio test. so taking a_k+1 / a_k you get after simplifying: 3/(M+1) (M+1 / M)^3 and when M goes to infinity, you get 0 so the series converges.