Question

when n->infinit, the limit of (1/n^3)(1^2+3^2+...+(2n-1)^2) =?

Answer 1

is (1^2+3^2+...+(2n-1)^2) in top or bottom?

Answer 2

top

Answer 3

(1^2+3^2+...+(2n-1)^2)/n^3 =? this looks better.

Answer 4

\[\lim_{n\to \infty } \, \frac{1^2+3^2+\text{...}+(2n-1)^2}{n^3}\]

Answer 5

yes . You use the tools to type this ?

Answer 6

yep, latex

Answer 7

Is latex the name of the tool? How can i get it? Can you give me a link?

Answer 8

no , you type the code, for example, I type this :\lim_{n\to \infty } \, \frac{1^2+3^2+\text{...}+(2n-1)^2}{n^3}

Answer 9

between \[ and

Answer 10

Wow..

Answer 11

That is right. but the result is still series

Answer 12

I am trying by squeez theorem..

Answer 13

The answer from the textbook is 4/3.

Answer 14

help me also pls

Answer 15

My summation was wrong,

\[\lim_{n\to \infty } \, \frac{\sum _{k=1}^n (-1+2 k)^2}{n^3}\]

\[\lim_{n\to \infty } \, \frac{\sum _{k=1}^n 4 (-1+2 k)}{3n^2}\]
\[\lim_{n\to \infty } \, \frac{\sum _{k=1}^n 8}{6}\]
\[\lim_{n\to \infty } \, \frac{8\sum _{k=1}^n 1}{6}\]

Answer 16

ignore my last two step :
\[\lim_{n\to \infty } \, \frac{\sum _{k=1}^n 8}{6n}\]
\[\lim_{n\to \infty } \, \frac{8\sum _{k=1}^n 1}{6n}\]
\[\lim_{n\to \infty } \, \frac{8n}{6n}\]
\[\lim_{n\to \infty } \, \frac{8n}{6n}=\frac{4}{3}\]

Answer 17

I see... Thank you so much . I learn a new skill for using L'Hospital's Rule. Thank you

Answer 18

you are welcome

Answer 19

I don't know the L'Hospital's Rule can use in that way before. I am so happy to learn it...HAHA

Answer 20

yeah they are usually best and easy way to determine limit. To do l'hospital , we need inderterminant of either 0/0 or infinity/infinity

Answer 21

Use l'hospital directly in the \[\sum_{?}^{?}\] is wonderful. I learn the skill.

Answer 22

Could you tell me more about the way you use l'hospital: I mean the way you take derivative directly in the ∑ .

Answer 23

I have another example:
Prove that :
\[if:\mathop {\lim }\limits_{n \to \infty } {a_n} = a\] then
\[\mathop {\lim }\limits_{n \to \infty } \frac{{{a_1} + 2{a_2} + ... + n{a_n}}}
{{{n^2}}} = \frac{a}
{2}\]

Answer 24

Can i use l'hospital rule like you did:
\[\mathop {\lim }\limits_{n \to \infty } \frac{{\sum\nolimits_{k = 1}^n {k{a_k}} }}
{{{n^2}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sum\nolimits_{k = 1}^n {{a_k}} }}
{{2n}} = \mathop {\lim }\limits_{n \to \infty } \frac{{an}}
{{2n}} = \frac{a}
{2}\]
Am i right? The new way of using l'hosptial rule is a little confused.