Question

Solve lim n-> (infinity) = 1/n^3 n(n +1)(2n + 1)/(6)

Answer 1

recheck the numerator, there is a gap between the first fraction.

Answer 2

it looks like this hold on \[\lim_{n \rightarrow infinity} = 1/n^3 = n(n + 1)(2n + 1)/6 \]

Answer 3

Do you mean this?
\[\lim_{n \rightarrow \infty} {1 \over n^3}\cdot{n(n+1)(2n + 1) \over 6}\]

Answer 4

yes

Answer 5

\[\lim_{n\rightarrow \infty} \frac{n(n+1)(2n+1)}{6n^3}\]
\[\frac{1}{6} \lim_{n\rightarrow \infty} \frac{(n + 1)(2n+1)}{n^2}\]
\[\frac{1}{6} \lim_{n \rightarrow \infty} \frac{2n^2 + 3n + 1}{n^2}\]
\[\frac{1}{6} \lim_{n \rightarrow \infty} \left(2 + \frac{3}{n} + \frac{1}{n^2}\right)\]

Which is \(\large {1 \over 6} * 2 = {1 \over 3}\)

Answer 6

okay

Answer 7

i'll try the second one of my self thanks!

Answer 8

Alright, if you get stuck you post it and I'll try to explain it :p

Answer 9

ok so for this one i got lim n -> infinity = 1/n^2 n(n + 1)/2 is a function with a numerator and denominator degree 2 to the limit is ration of the leading coefficients, namely 1/2

Answer 10

Yep :)

Answer 11

@Meepi so i got that one right ?

Answer 12

Yep

Answer 13

okay so the first one its 1/3 and this one 1/2

Answer 14

thanks!

Answer 15

no problem :D