Reimann sum related question from NBHM 2011(2.2)

Question

Evaluate: $$\lim_{n\to \infty}{\frac{1}{1+n^3}}+{\frac{4}{8+n^3}}+\dots+{\frac{n^2}{n^3+n^3}}.$$ I don't know how to proceed but the answer is $1/3\log (2)$.

Answer 1

It can be proven that the limit at infinity of the Riemann Sum is the exact area under the graph of $f$ from a to b. This limit has a special name and notation. It is called the definite integral.

Definition of Definite Integral If $f$ is a continuous function defined on [a,b], and if [a,b] is divided into n equal subintervals of width ∆x = $b − a \over n$ , and if $x_{k}$ = a + k ∆x is the right endpoint of subinterval k , then the definite integral of f from a to b is the number

$∫_{a}^{b}f(x)dx = lim_{n\to ∞}∑_{k=1}^{n}f(x_{k} )∆x$

ANSWER:

$lim_{n\to ∞} ({1 \over 1+n^3 }+{4 \over 8+n^3 }+...+{n^2 \over n^3+n^3 })=lim _{n\to ∞} \sum_{k=1}^{n}{k^2 \over k^3+n^3 } $

$lim _{n\to ∞} \sum_{k=1}^{n}{k^2 \over k^3+n^3 }= lim _{n\to ∞} \sum_{k=1}^{n}{{k^2\over n^3} \over ({k\over n})^3+1 } $

let ${k\over n}=x ==\Rightarrow lim _{n\to ∞}{1\over n}=dx \to 0 $

upper limit of $\space$ $x= lim _{n\to ∞}{n\over n} =1$

lower limit of $\space$ $x= lim _{n\to ∞}{1\over n}=0$

$lim _{n\to ∞} \sum_{k=1}^{n}{{x^2} \over x^3+1 } = \int _{0}^{1}{x^2dx\over x^3+1} $

=$[{ln(|x^3+1|)\over 3}+c]_{0}^{1}$

$=[{ln(2) \over 3 }-0]$

$={ln(2) \over 3 }$

Answer 2

Hint. We have that the sum can be written as $$\frac{1}{n}\sum_{k=1}^n\frac{(k/n)^2}{(k/n)^3+1}$$ Is it a Riemann sum?