Question

For positive integer n, defined


`\rm f(n) = n + \frac{16 + 5n - 3 {n}^(2) }{4n + 3 {n}^(2) } + \frac{32 + n - 3 {n}^(2) }{8n + {3n}^(2) } + \frac{48 - 3n - 3 {n}^(2) }{1 2n+3 {n}^(2) } + \cdots + \frac{25 {n} - 7n^(2) }{7 {n}^(2) }`
Then, the value of
`\displaystyle\rm\lim_(n \to \infty )` is equal to

Answer 1

Following the first term
`n`, the
`k`-th term
`\left(k\in\{1,2,3,\ldots,n\}\right)` in
`f(n)` is given by

`(16k + (9-4k)n - 3n^2)/(4kn + 3n^2)`

That is,

`k=1 \implies (16 + (9-4)n - 3n^2)/(4n + 3n^2) = (16 + 5n - 3n^2)/(4n + 3n^2)`

`k=2 \implies (16\cdot2 + (9-8)n - 3n^2)/(4\cdot2n + 3n^2) = (32 + n - 3n^2)/(8n + 3n^2)`

and so on, up to

`k=n \implies (16n + (9-4n)n - 3n^2)/(4n^2 + 3n^2) = (25n - 7n^2)/(7n^2)`

We then have

`\displaystyle f(n) = n + \sum_(k=1)^n (16k + (9-4k)n - 3n^2)/(4kn + 3n^2) \\\\ ~~~~~~~ = n + \frac1n \sum_(k=1)^n (16\frac kn + 9)/(4\frac kn + 3) - \frac1n \sum_(k=1)^n (4kn + 3n^2)/(4k + 3n) \\\\ ~~~~~~~ = n + \frac1n \sum_(k=1)^n (16\frac kn + 9)/(4\frac kn + 3)- \frac1n \sum_(k=1)^n n \\\\ ~~~~~~~ = \frac1n \sum_(k=1)^n (16\frac kn + 9)/(4\frac kn + 3)`

Now as
`n\to\infty`, the right side converges to a definite integral.

`\displaystyle \lim_(n\to\infty) f(n) = \lim_(n\to\infty) \frac1n \sum_(k=1)^n (16\frac kn + 9)/(4\frac kn + 3) \\\\ ~~~~~~~~~~~~~ \,= \int_0^1 (16x + 9)/(4x + 3) \, dx`

Wrap up by substituting
`x\mapsto\frac{x-3}4`.

`\displaystyle \lim_(n\to\infty) f(n) = \frac14 \int_0^1 (16x + 9)/(4x + 3) \, dx \\\\ ~~~~~~~~~~~~~\, = \frac14 \int_3^7 \frac{4x - 3}x \, dx \\\\ ~~~~~~~~~~~~~\, = \frac14 \int_3^7 \left(4 - \frac3x\right) \, dx \\\\ ~~~~~~~~~~~~~\, = \left. \frac14 (4x - 3\ln|x|) \right\vert_3^7\\\\ ~~~~~~~~~~~~~\, = \boxed{4 + \frac34 \ln\left(\frac37\right)}`