Question

How do you find a formula for the sum n terms `sum_(i=1)^n (1+i/n)(2/n)` and then find the limit as `n->oo`?

Answer 1

`sum_(i=1)^n (1+i/n)(2/n) = (3n+1)/n`

`lim_(n rarr oo)sum_(i=1)^n (1+i/n)(2/n) = 3`

Explanation:

Let `S_n = sum_(i=1)^n (1+i/n)(2/n)`
`:. S_n = sum_(i=1)^n (2/n+(2i)/n^2)`
`:. S_n = 2/n sum_(i=1)^n (1) + 2/n^2 sum_(i=1)^n (i)`

And using the standard results:

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

We have;

`S_n = 2/n(n)+2/n^2*1/2n(n+1)`
`:. S_n = 2+(n+1)/n`
`:. S_n = ((2n)+(n+1))/n`
`:. S_n = (3n+1)/n`

Now we examine the behaviour of `S_n` as `n rarr oo`.
We have;

`S_n = (3n+1)/n`
`:. S_n = 3+1/n`
`:. lim_(n rarr oo)S_n = lim_(n rarr oo) { 3+1/n }`
`:. lim_(n rarr oo)S_n = lim_(n rarr oo) (3) - lim_(n rarr oo)(1/n)`

And as `1/n rarr 0` as `n rarr oo`, we have;

`:. lim_(n rarr oo)S_n = 3`