Question

How do you find a formula for the sum n terms `Sigma(2i)/n(2/n)` and then find the limit as `n->oo`?

Answer 1

`sum_(i=1)^n (2i)/n(2/n) = 2 + 2/n`

`lim_(n rarr oo) sum_(i=1)^n (2i)/n(2/n) = 2`

Explanation:

We will need the standard results:
`sum_(r=1)^n r = 1/2n(n+1)`

Let

`S_n = sum_(i=1)^n (2i)/n(2/n)`
`:. S_n = 4/n^2sum_(i=1)^n i`
`:. S_n = 4/n^2{ 1/2n(n+1) }`
`:. S_n = 2/n(n+1)`
`:. S_n = 2 + 2/n`

Hence, `sum_(i=1)^n (2i)/n(2/n) = 2 + 2/n`

Consequently,

`lim_(n rarr oo) S_n = lim_(n rarr oo) (2 + 2/n)`
`lim_(n rarr oo) S_n = lim_(n rarr oo) (2) + lim_(n rarr oo) (2/n)`
`lim_(n rarr oo) S_n = 2 + 0`
`lim_(n rarr oo) S_n = 2`