Question

how do I go about evaluating sums

Answer 1

what do you man by that

Answer 2

\[\sum_{i=n}^{n}\left[ \frac{ i-n }{ n } \right]^\frac{ 1 }{ 2 }\left( \frac{ 1 }{ n } \right)\]

Answer 3

well then

Answer 4

Since it goes from \(n\) to \(n\), then just let \(i=n\) and you're done.

Answer 5

isn't it zero?

Answer 6

my bad
i = 1

Answer 7

oh ok so @wio is correct and i believe the answer is zero

Answer 8

correction
\[\sum_{i=1}^{n}\left[ \frac{ i-1 }{ n } \right]^\frac{ 1 }{ 2 }\left( \frac{ 1 }{ n } \right)\]