# Thread: Can't Figure Out Missing Step: lim[n->infty] (3/n^3) [n(n+1)(2n+1)]/[6] + (3/n)n

1. ## Can't Figure Out Missing Step: lim[n->infty] (3/n^3) [n(n+1)(2n+1)]/[6] + (3/n)n

The example is as follows:

. . .$\displaystyle \lim_{n \rightarrow \infty} \dfrac{3}{n^3}\, \dfrac{n\, (n\, +\, 1)\, (2n\, +\, 1)}{6}\, +\, \dfrac{3}{n}\, n$

. . .$\displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{2}\, \cdot\, \dfrac{n}{n}\, \cdot\, \left(\dfrac{n\, +\, 1}{n}\right)\, \left(\dfrac{2n\, +\, 1}{n}\right)\, +\, 3$

. . .$\displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{2}\, \cdot\, 1\, \left(1\, +\, \dfrac{1}{n}\right)\, \left(2\, +\, \dfrac{1}{n}\right)\, +\, 3$

. . . . . . . . . .$\displaystyle =\, \dfrac{1}{2}\, \cdot\, 1\, \cdot\, 1\, \cdot\, 2\, +\, 3\, =\, 4$

Where I start to get confused is on the second line as it appears that some intermediary steps were skipped for brevity. Normally, I would have solved the question like so:

. . .$\displaystyle \lim_{n \rightarrow \infty} \dfrac{3}{n^3}\, \dfrac{n\, (n\, +\, 1)\, (2n\, +\, 1)}{6}\, +\, \dfrac{3}{n}\, n$

. . .$\displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{2}\, \cdot\, \dfrac{1}{n^3}\, \cdot\, \left(2n^3\, +\, 3n^2\, +\, n\right)\, +\, 3$

. . .$\displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{2}\, \cdot\, \left(2\, +\, \dfrac{3}{n}\, +\, \dfrac{1}{n^2}\right)\, +\, 3$

. . . . . . . . . .$\displaystyle =\, \dfrac{1}{2}\, \cdot\, 2\, +\, 3\, =\, 4$

so trying to understand the question from a different perspective would be quite useful to me.

2. Your work is fine. All they did differently was to leave it in factored form, which saves work. They canceled the 3 and 6 to get 1/2; and they split the n^3 among the three factors, putting an n under each of the factors n, n+2, 2n+1.

3. Originally Posted by pipsydoodles
The example is as follows:

. . .$\displaystyle \lim_{n \rightarrow \infty} \dfrac{3}{n^3}\, \dfrac{n\, (n\, +\, 1)\, (2n\, +\, 1)}{6}\, +\, \dfrac{3}{n}\, n$

. . .$\displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{2}\, \cdot\, \dfrac{n}{n}\, \cdot\, \left(\dfrac{n\, +\, 1}{n}\right)\, \left(\dfrac{2n\, +\, 1}{n}\right)\, +\, 3$

. . .$\displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{2}\, \cdot\, 1\, \left(1\, +\, \dfrac{1}{n}\right)\, \left(2\, +\, \dfrac{1}{n}\right)\, +\, 3$

. . . . . . . . . .$\displaystyle =\, \dfrac{1}{2}\, \cdot\, 1\, \cdot\, 1\, \cdot\, 2\, +\, 3\, =\, 4$

Where I start to get confused is on the second line as it appears that some intermediary steps were skipped for brevity. Normally, I would have solved the question like so:

. . .$\displaystyle \lim_{n \rightarrow \infty} \dfrac{3}{n^3}\, \dfrac{n\, (n\, +\, 1)\, (2n\, +\, 1)}{6}\, +\, \dfrac{3}{n}\, n$

. . .$\displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{2}\, \cdot\, \dfrac{1}{n^3}\, \cdot\, \left(2n^3\, +\, 3n^2\, +\, n\right)\, +\, 3$

. . .$\displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{2}\, \cdot\, \left(2\, +\, \dfrac{3}{n}\, +\, \dfrac{1}{n^2}\right)\, +\, 3$

. . . . . . . . . .$\displaystyle =\, \dfrac{1}{2}\, \cdot\, 2\, +\, 3\, =\, 4$

so trying to understand the question from a different perspective would be quite useful to me.
They left things factored, and split things into multiple multiplied fractions. You multiplied things together, and split the one fraction into three. The result is the same, either way, and each is, I think, equally valid.

4. Awesome! Thank for the help.