Question

How do you use the summation formulas to rewrite the expression `Sigma (4j+3)/n^2` as j=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_(j=1)^n 1/n^2(4j+3) = (2n + 5)/n`

Explanation:

Let `S_n = sum_(j=1)^n 1/n^2(4j+3)`
`:. S_n = 1/n^2sum_(j=1)^n (4j+3)`
`:. S_n = 1/n^2{sum_(j=1)^n (4j) + sum_(j=1)^n (3)}`
`:. S_n = 1/n^2{4 sum_(j=1)^n (j) + sum_(j=1)^n (3)}`

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

We have;

`S_n = 1/n^2{4 1/2n(n+1) + 3n}`
`:. S_n = 1/n^2(2n(n+1) + 3n)`
`:. S_n = 1/n^2(2n^2+2n + 3n)`
`:. S_n = 1/n^2(2n^2 + 5n)`
`:. S_n = 1/n^2(2n + 5)n`
`:. S_n = (2n + 5)/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 2`; This is probably the conclusion of this question]

Now, `S_n = (2n + 5)/n`

`:. S_n = 2n/n + 5/n`
`:. S_n = 2 + 5/n`

And so,

`lim_(n rarr oo) S_n = lim_(n rarr oo) (2 + 5/n)`
`:. lim_(n rarr oo) S_n = lim_(n rarr oo) (2) + lim_(n rarr oo)(5/n)`
`:. lim_(n rarr oo) S_n = 2 + 5 lim_(n rarr oo)(1/n)`
`:. lim_(n rarr oo) S_n = 2`, `" as "(lim_(n rarr oo)(1/n) =0)`
Which confirms our assumption!