$\lim_{n→∞} \frac{1^3 +4^3 +7^3 + ... + (3n-2)^3}{[(1+4+7+...+(3n-2)]^2}$
I'm mostly confused because I'm not sure what the raising to the second power in the denominator means. Does this mean that after I sum all of the terms of (3n-2) from n to infinity I then multiply that by itself?
Do I need to use Riemann sums here?
I tried to do it with the squeeze theorem but ended up getting infinity on the right side each time.
(I've looked around in "similar questions" to see if this has been asked before and as far as I've seen I couldn't find anything that helps me with this in previously asked questions.)