Crazy Limits 2

Calculus Level 3

$\Large \lim_{n \to \infty} \frac{1^2+2^2+3^2......+n^2}{n^3} = \, ?$

1 solution

First Last
Jan 11, 2016

The Numerator is clearly $\sum_{i=1}^n (i^2)$ which is equal to $\frac{n(n+1)(2n+1)}{6}$

Then $\frac{\frac{n(n+1)(2n+1)}{6}}{n^3}$ reduces to:

$\frac{1}{3} + \frac{1}{2n} + \frac {1}{6n^2}$

So as n --> ∞, only $\boxed{\frac{1}{3}}$ matters!

I just missed 6 in the denominator of submission of n^2

Raghav Rathi - 4 years, 10 months ago