Question

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 1

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.

Answer 2

i am thinking. maybe can prove it directly?

Answer 3

we can use l'hospital

Answer 4

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

Answer 5

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

Answer 6

divide by n^2 take the limit in your head

Answer 7

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

Answer 8

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

Answer 9

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...

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

it is a/2

Answer 14

what did i do wrong then

Answer 15

you caont do what you did above

Answer 16

wait don't answer that

Answer 17

i see

Answer 18

its because i'm forgeting the terms before na_n

Answer 19

yes

Answer 20

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

Answer 21

yes

Answer 22

and so on...

Answer 23

yes

Answer 24

lol

Answer 25

i will let you prove it

Answer 26

because i know you want to badly

Answer 27

lol

Answer 28

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

Answer 29

how do you factor out an a?

Answer 30

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

Answer 31

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

Answer 32

wheres that?

Answer 33

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

Answer 34

an epsilon based proof is the way to go I believe

Answer 35

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?