Cubic Limit

Calculus Level pending

$\large\lim \limits_{n\to \infty }\dfrac{1}{n^4}\sum \limits^{2n - 1}_{j=0} j^3 = \ ?$

1 8 4 16

1 solution

Chew-Seong Cheong
Mar 29, 2017

$\begin{aligned} L & = \lim_{n \to \infty} \frac 1{n^4} \sum_{j=0}^{2n-1} j^3 \\ & = \lim_{n \to \infty} \frac 1{n^4} \left( \frac {(2n-1)(2n)}2 \right) ^2 \\ & = \lim_{n \to \infty} \frac {4n^2(4n^2+4n+1)}{4n^4} \\ & = \lim_{n \to \infty} \frac {4n^2+4n+1)}{n^2} \\ & = \lim_{n \to \infty} 4 + \frac 4n + \frac 1{n^2} \\ & = \boxed{4} \end{aligned}$

Nice, did the same only struggled because a accidentally forgot the square at first...

Peter van der Linden - 4 years, 2 months ago

Sir you made a mistake here

${\left( 2n-1 \right)}^2 = 4n^2 - 4n + 1$

Tapas Mazumdar - 4 years, 2 months ago

Thanks. I have changed it.

Chew-Seong Cheong - 4 years, 2 months ago

Cubic Limit

1 solution

0 pending reports