Working on sequence, possibly recursive

Question

I am working on this problem which asks to find if the sequence converges or not and if so the value it converges to. I am not sure how to deal with this type of question, but I feel like it may be a recursive relation. It is $a_n=\dfrac{1^2}{n^3}+\dfrac{2^2}{n^3}+ \cdots +\dfrac{n^2}{n^3}$

What I have tried was calculating the first couple terms, $a_1=1$, $a_2=0.625$, $a_3=0.5185$ etc.I also tried writing it as $\dfrac{1^2+2^2+..+n^2}{n^3}$, but I am just really lost on where to go with this.

Thanks all

Answer 1

First, note that $1^2+2^2+\cdots+n^2 = \frac{n(n+1)(2n+1)}{6}$. Then: \begin{align} \frac{1^2}{n^3} + \cdots + \frac{n^2}{n^3} &= \frac{\frac{n(n+1)(2n+1)}{6}}{n^3}\\ &\underset{n\to\infty}{=} \frac{2n^3}{6n^3}\\ &= \frac{1}{3} \end{align}

Answer 2

Note that $a_n=\dfrac{n(n+1)(2n+1)}{6n^3}=\dfrac{n^3(1+1/n)(2+1/n)}{6n^3} \rightarrow \dfrac{1}{3}$

Answer 3

If you don't know the closed form for the sum of squares, you can consider $\displaystyle \dfrac{1^2}{n^3} + \ldots + \dfrac{n^2}{n^3}$ as a Riemann sum for the integral $\displaystyle \int_0^1 x^2\; dx$.