Question

lim (1/n^2 + 2/n^2 + ... + n-1/n^2) . n->infinity . n is Z .

Answer 1

james help me

Answer 2

denominators are all
\[n^2\]? if so you can add these up.

Answer 3

get
\[\frac{1+2+...+(n-1)}{n^2}\] and the numerator sums to
\[\frac{n(n-1)}{2}\]
so you should be in good shape to find the limit

Answer 4

btw,

n\in \mathbb Z
translates as
\[ n \in \mathbb Z\]

Answer 5

Yes , I know that . Thanks !

Answer 6

yw