Limits (1 Viewer)

InteGrand · Dec 26, 2016

nrumble42 said:
View attachment 33659

Answer: 1/4.

leehuan · Dec 26, 2016

No idea why that question was given without any help whatsoever, but you need to prove that 1^3+2^3+...+n^3 = 1/4 n^2 (n+1)^2

InteGrand · Dec 26, 2016

$\noindent Note that the given limit is just the limit of a Riemann sum for $\int_{0}^{1}x^{3}\,\mathrm{d}x$, so has value $\frac{1}{4}$.$

$\noindent Just in general we have the following results. Fix a positive integer $d$ and let $S_{d}(n) = 1^{d}+2^{d} + \cdots + n^{d}$, for positive integers $n$. Then:$

$\noindent 1) $S_{d}(n)$ is a degree $d+1$ polynomial in $n$ with leading coeffficient $\frac{1}{d+1}$ (This implies the next result.)$

$\noindent 2) $\frac{S_{d}(n)}{n^{d+1}} \to \frac{1}{d+1}$ as $n\to \infty$$

$\noindent 3) $\frac{1}{d+1} n^{d+1} < S_{d}(n) < \frac{1}{d+1}(n+1)^{d+1}$.$

$\noindent To prove 3) (which also implies 2) by the squeeze law and answers the question at hand), consider inequalities from upper and lower Riemann sums for $\int_{0}^{1}x^{d}\,\mathrm{d}x$. Of course $\frac{S_{d}(n)}{n^{d+1}}$ is also a Riemann sum for $\int_{0}^{1}x^{d}\,\mathrm{d}x$, so also has limit $\frac{1}{d+1}$.$

Drongoski · Dec 26, 2016

nrumble42 said:
View attachment 33659

= $\lim _ {n \to \infty }\frac {[\frac {n(n+1)}{2}]^2} {n^4} = \frac {1}{4}\lim _{n\to \infty} {[1(1 + \frac {1}{n})]^2} = \frac {1}{4} [1+0]^2 = \frac {1}{4}$

Note- Well-known formula:

$1^3 + 2^3 + 3^3 + \cdots + n^3 = [1+2+3 + \cdots + n]^2 = [\frac {n}{2}(n+1)]^2$

Limits (1 Viewer)

nrumble42

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