English isn't the language I'm taught maths in, so I'm not sure how much sense the subject question makes, but never mind that. The question is, how can I calculate the sum of a sequence that goes 1^3 + 2^3 + 3^3 + 4^3 + ... + n^3? That is, how can I express the sum as a function of n? My book only contains formulas for the sum of arithmetic and geometric sequences, which I doubt this sequences qualifies as, and I haven't had any luck searching the web. Help appreciated. Edit: Forget it, I figured out how to solve the problem.
The series doesn't converge, so the sum is infinity. For a series Sum (1 -> infinity) a<sub>n</sub> to converge, a necessary, but not sufficient, condition is that a<sub>n</sub> tends to zero as n tends to infinity. In your example a<sub>n</sub> = n<sup>3</sup>, which tends to infinity as n tends to infinity. Hence the series does not converge.
James, is the fact that the sum of the infinite series is infinity relevant to the problem? Didn't he just want the sum of n terms?
The answer to this is quite simple. This is known as a genmetric sequence due to the constant intervals of each mathematical term. 1, 2, 3, 4... is an arathmatic sequence because there is no constant interval from one number to the next. I'm sure you know this. The qutoed answer is incorrect n(n+1) / 2 would yeild the sum from 1 to a set end, this however is only correct for an arathmatic sequence. Anyway to solve this problem you will have to use a simation of n: E = simation E (n+1)^3 n=0
E = simation Althought the Greek capital sigma, Σ, most closely resembles the letter 'E' in the Roman alphabet, it is actually the Greek letter 'S'. There is a summation symbol in HTML, it is a larger sigma, it looks like this: ∑
S. Dalai: The sequence orginally given 1+ 2^n+ 3^n+ ... , for fixed n, is NOT a geometric series. A geometric series is of the form 1+ r+ r^2+ r^3+ ... in which there is a constant ratio between terms. The quoted answer (n(n+1)/2)^2 (not n(n+1)/2) IS correct for the case of 1+ 2^3+ 3^3+...+ n^3 .
