Question

Evaluate lim(N->infinity) of E(k=1 to N) (2k/N^2)

Answer 1

is this the question?
$$
\Large {
\lim_{n\to \infty } \sum_{k=1}^{n}~\frac{2k}{n^2}
}

$$

Answer 2

yes, thanks

Answer 3

not sure how to begin

Answer 4

we can factor out 2/n^2
$$
\Large {
\lim_{n\to \infty } \sum_{k=1}^{n}~\frac{2k}{n^2}

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

}

$$

Answer 5

now we know the sum of k , starting from k=1 to n.
i can show you that separately

Answer 6

$$
\Large {
\lim_{n\to \infty } \sum_{k=1}^{n}~\frac{2k}{n^2}

\\\lim_{n\to \infty } \frac{2}{n^2} \cdot \sum_{k=1}^{n} k
\\\lim_{n\to \infty } \frac{2}{n^2} \cdot \frac{n(n+1)}{2}


}

$$

Answer 7

now try to find the limit of that

Answer 8

can you tell my why it was possible to factor out all but k initially?

Answer 9

k stands for k=1,2,3,... up to n
so we cant take that out

Answer 10

but we can take out n^2

Answer 11

so, can sigma be distributed?

Answer 12

sorry i dont understand your question

Answer 13

with respect to sigma, k is a variable
2/n^2 is a constant

Answer 14

ok, think I understand

Answer 15

cancelling the twos, we end up with (n^2 + n)/N^2 or (n + 1) / n

Answer 16

and using the leading coefficient, I come up with an answer of 1, is that right?