Question

Question in comment

Answer 1

okay

Answer 2

Let me try:

\(\dfrac{n+1}{n^2}, \dfrac{n+2}{n^2},\dfrac{n+3}{n^2},\cdots,\dfrac{n+(n^2-n)}{n^2}\)

So for that remaining part, there is \(n^2-n\) sub-intervals. Right?

Answer 3

yes

Answer 4

and how many points under the form 1/n in that part?

Answer 5

that bothers me a lot. Since if n =2, then \(\dfrac{2+2}{2^2}=1\) and then we are done right after 2, how can we have n^2-n sub intervals?

Answer 6

Exactly what do you mean?

\(2^2-2 = 2\)

Answer 7

But after that point n=3, we have \(\dfrac{5}{9}\) it is in [1/n,1] ha!!!

Answer 8

\[
3^2-3=6
\]