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

Can I use L'hospital Rule Like that: \[\mathop {\lim }\limits_{n \to \infty } \frac{{\sum\nolimits_{k = 1}^n {k{a_k}} }} {{{n^2}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sum\nolimits_{k = 1}^n {{a_k}} }} {{2n}} = \mathop {\lim }\limits_{n \to \infty } \frac{{an}} {{2n}} = \frac{a} {2}\] Is it a rigorous way to prove that? Please tell me your thought. It will be a appreciated.

i am thinking. maybe can prove it directly?

we can use l'hospital

I don't know how to do that, I want to use the precise definiition of series limit.

you need epsilon, N proof? because it looks like you should get \[a\sum_{k=1}^nk=\frac{a(n(n+1)}{2}\]

divide by n^2 take the limit in your head

In fact, it is a question from mathematical analysis. So yes, epsilon, N proof will be best.

\[\lim_{n \rightarrow \infty}\frac{a_1+a_2+\cdot \cdot \cdot n a_n}{n^2} \cdot \frac{\frac{1}{n^2}}{\frac{1}{n^2}}\]\[=\lim_{n \rightarrow \infty}\frac{\frac{a_1}{n^2}+\frac{a_2}{n^2}+ \cdot \cdot \cdot +\frac{n a_n}{n^2}}{\frac{n^2}{n^2}}\] \[=\lim_{n \rightarrow \infty}\frac{0+0+\cdot \cdot \cdot +\frac{1}{n}a_n}{1}=\lim_{n \rightarrow \infty}\frac{a_n}{n}=0\] ? how can we show its \[\frac{a}{2}\]

The answer from textbook is a/2. And I saw a guy use l'hospital rule in that way, so i just copy his way....I hope It work. But The way use l'hospital rule like that is really confused... AND If you can give the " epsilon-N " proof, that will be best...

Take your time. I have go to work now. See you.

i'm saying i can't prove that because it doesn't = a/2 i get 0 see above

and the way i used above wasn't l'hospital

it is a/2

what did i do wrong then

you caont do what you did above

wait don't answer that

i see

its because i'm forgeting the terms before na_n

yes

like (n-1)a_{n-1}

yes

and so on...

yes

lol

i will let you prove it

because i know you want to badly

lol

why isn't it \[\lim_{n\rightarrow \infty}\frac{an(n+1)}{2n^2}=\frac{a}{2}\]?

how do you factor out an a?

that is what it looks like you are doing...replacing a_n with a

the L'Hospitals rule use by prost in his first post is incorrect too

wheres that?

this ... \[\mathop {\lim }\limits_{n \to \infty } \frac{{\sum\nolimits_{k = 1}^n {k{a_k}} }} {{{n^2}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sum\nolimits_{k = 1}^n {{a_k}} }} {{2n}} = \mathop {\lim }\limits_{n \to \infty } \frac{{an}} {{2n}} = \frac{a} {2}\]

an epsilon based proof is the way to go I believe

I am back.. Hard Day.. Does anyone heard Stolz Theorem? Use it, it said: \[\mathop {\lim }\limits_{n \to \infty } \frac{{{a_1} + 2{a_2} + ... + n{a_n}}} {{{n^2}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n{a_n}}} {{{n^2} - {{\left( {n - 1} \right)}^2}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{an}} {{2n - 1}} = \frac{a} {2}\] and foreget my first proof above. It makes people confused and make no sense. The Stolz Theorem is very useful. But i still want the " epsilon-N " proof. Can anyone prove it?

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