How do I find a solution to this this finite series? $ \frac{1}{n^4} \sum_{i=1}^{n} \left({i^3}\right) $

Question

How do I find a solution to this this finite series? Any help would be greatly appreciated.

$$ \frac{1}{n^4} \sum_{i=1}^{n} \left({i^3}\right) $$

Just check out or google the formula on how to add cubes of consecutive integers and you are through. — Devarsh Ruparelia, Apr 24 '15 at 18:01
I gave a derivation of the formulas others have quoted here: http://math.stackexchange.com/questions/1194072/where-do-summation-formulas-come-from/1194116#1194116 — Ian, Apr 24 '15 at 18:08
Well, TIL of the "trivial answer converted to comment" function. I'm not sure I like it in this case... — Ian, Apr 24 '15 at 18:09

score 2 · Answer 1 · answered Apr 24 '15 at 18:05

2

Hint: $$\sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2.$$

answered Apr 24 '15 at 18:05

Ebearr

866

score 1 · Answer 2 · answered Apr 24 '15 at 18:04

1

You may want to know Faulhaber's formula. $$\sum_{i=1}^{n}i^3=\left(\frac{n(n+1)}{2}\right)^2.$$

answered Apr 24 '15 at 18:04

mathlove

139,939

And then dividing by $n^4$ turns the $n^2$ into $n^{-2}$ – Simpson17866 Apr 24 '15 at 18:06

Elaqqad · Answer 3 · 2015-04-24T18:15:54.370

1

If you're interested by the limit, we can also use Riemann sums: $$\frac{1}{n^4}\sum_{i=1}^n i^3 =\frac{1}{n}\sum_{i=1}^n \left(\frac i n\right)^3 \xrightarrow{n\to \infty} \int_{0}^1x^3dx=\frac{1}{4}$$

This method can be generalized for higher powers.

edited Apr 24 '15 at 18:15

answered Apr 24 '15 at 18:09

Elaqqad

13,725

How do I find a solution to this this finite series? $ \frac{1}{n^4} \sum_{i=1}^{n} \left({i^3}\right) $

3 Answers3