Question

find the formula for the sum of n terms then use the formula to find the limit as n -> infinity

Answer 1

\[\lim_{n \rightarrow \infty} \sum_{i=1}^{n}\frac{ 1 }{ n^3 }(i-1)^2\]
setting (i-1) to u, so this would then be
\[\lim_{n \rightarrow \infty} \sum_{i=1-1}^{n-1}\frac{ 1 }{ n^3 }(u)^2\]

Answer 2

WHAT THE HICKIE IS INFINITY - 1?

Answer 3

Infinity - (real number) = infinity

Answer 4

@sourwing did I set the sigma correctly?

Answer 5

or instead of i = 1 it would be u = 1-1 \[\lim_{n \rightarrow \infty} \sum_{u=1}^{n}\frac{ 1 }{ n^3 }(i-1)^2\]

Answer 6

\[\lim_{n \rightarrow \infty} \sum_{u=1-1}^{n-1}\frac{ 1 }{ n^3 }(i-1)^2\]

Answer 7

well, again the increment is i, not n. So n is like a constant

Answer 8

\[\lim_{n \rightarrow \infty} \sum_{u=1-1}^{n-1}\frac{ 1 }{ n^3 }(u)^2\]

Answer 9

so it's not even correct to touch n really?

Answer 10

pull n^3 out of sigma,
the sum is just sum of squares of first (n-1) natural numbers

Answer 11

\(\large \lim \limits_{n \rightarrow \infty} \sum \limits_{u=1-1}^{n-1}\frac{ 1 }{ n^3 }(u)^2 = \lim \limits_{n \rightarrow \infty} \frac{ 1 }{ n^3 }\sum \limits_{u=0}^{n-1}(u)^2\)

Answer 12

\[\lim_{n \rightarrow \infty} \frac{ 1 }{ n^3 }\sum_{i=1}^{\infty }(i-1)^2\]

Answer 13

if you let u = i-1,
when i = 1, u = 0
when i = n, u = n-1

so, \[\lim n \rightarrow \infty \sum_{u=0}^{n-1} u^2\]

which is the same as lim n rightarrow infty sum_{u=1}^{n-1} u^2

\[(1/6) u(u+1)(2u+1)\]

Answer 14

yes both are same,
u or i... dont mess wid limit yet... .

Answer 15

@nincompoop
Forget the limit and solve sigma first.
You can apply limit later.

Answer 16

plug in n-1 for u
(1/6) (n-1) (n-1+1) (2(n-1)+1)

Answer 17

\[\lim_{n \rightarrow \infty} \frac{ 1 }{ n^3 }\sum_{u=0}^{n-1 }(u)^2\]

Answer 18

(1/6) (n-1)(n-1+1) (2(n-1)+1) / n^3 has limit of 1/2 as n->inf

Answer 19

I see what you did there @sourwing

Answer 20

you plugged -1 on all n's then proceeded with summation formulas

Answer 21

yep

Answer 22

Did i say 1/2? I meant 1/3
:DD

Answer 23

I was hoping you would correct that because I remember you said it's the fraction value that it approaches or something of that sort

Answer 24

yeah ^^

Answer 25

I don't even need to evaluate different values that saves a lot of time