Question

So remember that
\[ \lim_{ n \rightarrow \infty} \frac{1}{n^2}=0 .\]
Why is it that you can't say the limit for the following is 0?
\[\lim_{n \rightarrow \infty } \frac{1+2+ \cdots +n}{n^2}=0+0+0+ \cdots +0=0\] but this limit is suppose to be 1/2
like this is probably a stupid question

Answer 1

How come now you have:\[
\lim_{n\to 0}\frac 1{n^2} =0
\]

Answer 2

oops that was suppose to be infinity there at that one spot

Answer 3

no this just a question i have been thinking about for awhile

Answer 4

The reason you can't do it is because the rule for limits over addition can work a finite number of times, but not an infinite number of times.

Answer 5

Recall that \(1+2+\cdots+n = \dfrac{n(n+1)}{2}\).

So, \(\displaystyle \lim_{n\to 0}\frac{1+2+\cdots n}{n^2 } = \lim_{n\to0} \frac{n(n+1)}{2n^2} = \frac{1}{2}\lim_{n\to 0} \left(1+\frac{1}{n}\right)\).

However, this doesn't have a limit as \(x\to0\). It has a limit of \(\dfrac{1}{2}\) as \(x\to\infty\), though.

Is that what you meant? :-)

Answer 6

I mean as \(n\to 0\), not \(x\to 0\).

Answer 7

...and the same for \(n\to\infty\) instead of \(x\to\infty\) (I'm getting my variables mixed up for some reason)

Answer 8

"The Limit Does not Exist" - Mean Girls

Answer 9

\[\lim_{n \rightarrow \infty}(\frac{1+2+\cdots +n}{n^2})=\lim_{n \rightarrow \infty}(\frac{1}{n^2})+\lim_{n \rightarrow \infty}(\frac{2}{n^2})+\cdots +\lim_{n \rightarrow \infty}(\frac{n}{n^2}) \\=0+0+\cdots+0=0\]
so by what wio said we can't do this

Answer 10

And I know that 1+2+...+n=n(n+1)/2

Answer 11

Basically, I would suggest that: \[

\lim_{n\to \infty}\frac{\sum_{k=1}^n k}{n^2} \neq \sum_{k=1}^n\lim_{n\to \infty}\frac k{n^2}
\]

Answer 12

Well, I am being a bit presumptuous. Basically the problem lies somewhere before the first step and the second step.

Answer 13

Ok thanks you guys
I guess I just forgot that we can only do that addition thing a finite number of times
That makes sense wio

Answer 14

.....99999999999999999999999999999
+1
----------------------------------------
0

Answer 15

for some reason ur infinite additions reminds of that hmm

Answer 16

so smart

Answer 17

It's a good example of a case where induction doesn't work to infinity.

Answer 18

I blame me forgetting this because of that feral cat that I mistook as a rat.
I went outside and I heard something in a box. Ran back inside. But it was just a kitty. Small and cute.

Answer 19

I was kidding. I just forgot. That is all.

Answer 20

But there really was a kitty outside today that I thought was a rat.

Answer 21

at some point it will be recursive like
1/n^2+2/n^2+3/n^2+....+(n-3)/n^2+(n-2)/n^2+(n-1)/n^2+n/n^2
1/n^2+2/n^2+3/n^2+....-3/n-2/n-2/n -1/n +n/n
so u would have
sum i /n^2 as n goes to infinity - sum i /n as n goes to infinity + 1 :O
that leads to nothing :3 but just away to say not all terms goes to 0