Question

Find a formula for the sum of n terms. Use the formula to find the limit as n → [infinity]. lim n → [infinity] n 2 + i n 8 n i = 1

Answer 1

Complete Question

Find a formula for the sum of n terms.
`\sum\limits_(i=1)^n ( 8 + (i)/(n) )((2)/(n) )`

Use the formula to find the limit as
`n \to \infty`

Answer:

`K_n = (n + 73 )/(n)`

`\lim_(n \to \infty) K_n = 1`

Explanation:

So let assume that

`K_n = \sum\limits_(i=1)^n ( 8 + (i)/(n) )((2)/(n) )`

=>
`K_n = \sum\limits_(i=1)^n ( (16)/(n) + (2i)/(n^2) )`

=>
`K_n = (2)/(n) \sum\limits_(i=1)^n (8) + (2)/(n^2) \sum\limits_(i=1)^n(i)`

Generally

`\sum\limits_(i=1)^n (k) = (1)/(2) n (n + 1)`

So

`\sum\limits_(i=1)^n (8) = (1)/(2) * 8* (8 + 1)`

`\sum\limits_(i=1)^n (8) = 36`

`K_n = (2)/(n) \sum\limits_(i=1)^n (8) + (2)/(n^2) \sum\limits_(i=1)^n(i)`

and

`\sum\limits_(i=1)^n (i) = (1)/(2) n (n + 1)`

Therefore

`K_n = (72)/(n) + (2)/(n^2) * (1)/(2) n (n + 1 )`

`K_n = (72)/(n) + (1)/(n) (n + 1 )`

`K_n = (72)/(n) + 1 + (1)/(n)`

`K_n = (72 + 1 + n )/(n)`

`K_n = (n + 73 )/(n)`

Now
`\lim_(n \to \infty) K_n = \lim_(n \to \infty) [(n + 73 )/(n) ]`

=>
`\lim_(n \to \infty) [(n + 73 )/(n) ] = \lim_(n \to \infty) [(n)/(n) + (73 )/(n) ]`

=>
`\lim_(n \to \infty) [(n + 73 )/(n) ] = \lim_(n \to \infty) [1 + (73 )/(n) ]`

=>
`\lim_(n \to \infty) [(n + 73 )/(n) ] = \lim_(n \to \infty) [1 ] + \lim_(n \to \infty) [(73 )/(n) ]`

=>
`\lim_(n \to \infty) [(n + 73 )/(n) ] = 1 + 0`

Therefore

`\lim_(n \to \infty) K_n = 1`