Find the limit as n approaches infinity for n sigma k=0 (k/n^2)

Question

solomonzelman · Answer

$$\lim_{x \rightarrow ∞}~~ \sum_{k=0}^{n}(k/n^2)$$ Like this ?

anonymous · Answer

yes, thanks

solomonzelman · Answer

I mean it should be n not x.$$\lim_{n \rightarrow ∞} ~\sum_{k=0}^{n}k/n^2$$

solomonzelman · Answer

you are trying to find it when k=0 ?

anonymous · Answer

yes, and it is approaching infinite. yes, k=0

solomonzelman · Answer

Well, just using logic (sorry for late reply, I was finishing a chess game), when k=0, wouldn't you get the output of zero (unless n=0)  ?
So whatever n equals if it is not equal to zero, when k=0, you will get a zero.

anonymous · Answer

so the limit is zero?

solomonzelman · Answer

wouldn't really make sense to put a point that is (∞,0)  Not an interval: a POINT !
Saying that it is limited at the coordinate (∞,0) ??
I would just say that just the function stops at x=0, like it is a removable discontinuity (If I am naming it correctly).

anonymous · Answer

i simplified the equation to $$\lim_{n \rightarrow infinite} 1/n^2  [n(n+1)/2] but im \not sure if that is correct$$

anonymous · Answer

i have no clue what the answer is

solomonzelman · Answer

Yes when k=1, it would be different

solomonzelman · Answer

When k=1, we know that
n^2 → ∞  and therefore,
the close n^2 approaches ∞, the smaller the  ` 1 /n^2` is going to be.

Saying,  n^2 →∞,  then  1/n² → 0