# Simple limit problem

• Apr 23rd 2011, 11:25 PM
sedam7
Simple limit problem
hello,
I'm trying to recall how to do this and I got stuck big time... I would be most grateful for any help :D

problem is :

I'll fix it in latex as soon it work with limits :D

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

and I recall that "limit of sum" is equal to "sum of limits"... something like ::
if is \lim_{n \to \infty} f(x) = A and \lim_{n \to \infty} f(x) = B than

\lim_{n\to \infty} ( f(x) +- g(x) ) = lim_{n\to \infty} f(x) +- lim_{n\to \infty} g(x)

and if I apply that to problem that I have, result would be zero but book says something else :( :( so now I'm in big trouble....

I also have more or less the same another problem but it's to "ugly" to write it without LATEX so...

what do I do wrong ?
• Apr 23rd 2011, 11:39 PM
Prove It
It will probably help if you realise that since there's a common denominator, you can write

1/n^2 + 2/n^2 + 3/n^2 + ... + (n - 1)/n^2 = [1 + 2 + 3 + ... + (n-1)]/n^2 = [n(n-1)/2]/n^2 = (n^2 - n)/2n^2 = 1/2 - 1/2n.

I'm sure you can take the limit now.
• Apr 23rd 2011, 11:52 PM
sedam7
Thank you :D very very much :D:D:D

I'm just curios now, why I can't apply way that I was going...

and if I have sequence with just odd numbers .... (1+3+5+....+(2n-1) )/(n+1) I would have what ? I think I don't really see from what do u conclude that it's [n(n-1)/2]/n^2 ...
hehehe or I'm missing basic stuff here :(
• Apr 23rd 2011, 11:57 PM
Prove It
The top is still an arithmetic series, so you can use S = N(t_1 + t_N)/2 to simplify...
• Apr 24th 2011, 12:03 AM
sedam7
ah... OK :D
meaning that every time that I have infinite sequence like 1+2+3+... +n :D I have to look at it like series :D and go to nth partial sum :D
Thanks :D
• Apr 24th 2011, 05:34 AM
yeKciM
or simple use Stolz th that says ....

if you are given two sequences (x_n) and (y_n) and if
1° lim_(n->\infty) (x_n) = +\infty
2° y_(n+1) >= y_n

than if exist

\lim_{n\to \infty} \frac {x_{n+1}-x_n }{y_{n+1}-y_n }

and it's equal

\lim_{n\to \infty} \frac {x_n}{y_n}=\lim_{n\to \infty} \frac {x_{n+1}-x_n }{y_{n+1}-y_n }