Question

How do you use the summation formulas to rewrite the expression `Sigma (2i+1)/n^2` as i=1 to n without the summation notation and then use the result to find the sum for n=10, 100, 1000, and 10000?

Answer 1

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

Explanation:

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

And using the standard results:
`sum_(r=1)^n r = 1/2n(n+1)`

We have;

`\ \ \ \ \ S_n = 1/n^2{2*1/2n(n+1) + n}`
`:. S_n = 1/n^2{n(n+1) + n}`
`:. S_n = 1/n^2{n^2+n + n}`
`:. S_n = 1/n^2{n^2+2n}`
`:. S_n = 1/n^2{n(n+2)}`
`:. S_n = (n+2)/n`

And this has been calculated using Excel for `n=10, 100, 1000, 10000`

What happens as `n rarr oo`?

[ NB As an additional task we could possibly conclude that as `n rarr oo` then `S_n rarr 1`; This is probably the conclusion of this question]

Now, `S_n = (n+2)/n`

`:. S_n = 1+2/n`

And so,

`lim_(n rarr oo) S_n = lim_(n rarr oo) (1+2/n)`
`:. lim_(n rarr oo) S_n = lim_(n rarr oo) (1) + 2lim_(n rarr oo)(1/n)`
`:. lim_(n rarr oo) S_n = 1+0`
`:. lim_(n rarr oo) S_n = 1`
Which confirms our assumption!