Why does \frac{b^3}{n^3} \frac{n(n+1)(2n+1)}{6} approach b^3/3 when n is large?

[Artusartos]

I don't really understand why $\frac{b^3}{n^3} \frac{n(n+1)(2n+1)}{6}$ is close to b^3/3 if n is very large...can anybody explain this to me?

I attached the problem...

Thanks in advance

 

[Robert1986]

If $n$ is big, then $\frac{n(n+1)(2n+1)}{6} \simeq 2n^3/6$ (in the sense that as $n \to \infty$ we have $\frac{n(n+1)(2n+1)}{6} \to \frac{2n^3}{6}$).

 

[Mark44]

Expand n(n + 1)(2n + 1), and then factor out n³. Finally, take the limit of your expression as n gets large.

 

[Ray Vickson]

Expand n(n + 1)(2n + 1), and then factor out n³. Finally, take the limit of your expression as n gets large.

In other words, write
$$\frac{n(n+1)(2n+1)}{n^3} = <br /> \frac{2 n^3}{n^3} \left(1 + \frac{1}{n}\right) \left( 1 + \frac{1}{2n} \right).$$

RGV

 

[Artusartos]

In other words, write
$$\frac{n(n+1)(2n+1)}{n^3} = <br /> \frac{2 n^3}{n^3} \left(1 + \frac{1}{n}\right) \left( 1 + \frac{1}{2n} \right).$$

RGV

Thanks :)