<s>[INST] Finding the value of $\mathop {\lim }\limits_{n \to \infty } f\left( n \right)$ of a series having the value "n" in both numerator and denominator If $f\left( n \right) = n + \frac{{16 + 5n - 3{n^2}}}{{4n + 3{n^2}}} + \frac{{32 + n - 3{n^2}}}{{8n + 3{n^2}}} + \frac{{48 - 3n - 3{n^2}}}{{12n + 3{n^2}}} + .. + \frac{{25 - 7{n^2}}}{{7{n^2}}}$. Then find the value of $\mathop {\lim }\limits_{n \to \infty } f\left( n \right) = \_\_\_$
My approach is as follow
$f\left( n \right) = n + \frac{{16 + 5n - 3{n^2}}}{{4n + 3{n^2}}} + \frac{{32 + n - 3{n^2}}}{{8n + 3{n^2}}} + \frac{{48 - 3n - 3{n^2}}}{{12n + 3{n^2}}} + .. + \frac{{25 - 7{n^2}}}{{7{n^2}}}$
$16 + \left( {n' - 2} \right)16 = 16n' - 16::\left( {5 + \left( {n' - 2} \right)\left( { - 4} \right)} \right) = 13 - 4n'::4 + \left( {n - 2} \right)4 = 4n' - 4$
${T_{n'}} = \frac{{16n' - 16 + 13n - 4n'n - 3{n^2}}}{{4n'n - 4n + 3{n^2}}}$
How will I proceed form here
 [/INST] $f(n) =  \sum_{r =1}^n (1 +  \frac{16r + (9-4r)n -3n^2}{4rn + 3n^2}) $
$  =  \sum_{r =1}^n ( \frac{16r +9n}{4rn +3n^2} ) \\ $
$ \lim_{n\to\infty} f(n)  = \lim_{n\to\infty} \frac{1}{n} ( \sum_{r =1}^n ( \frac{ 16(\frac{r}{n}) +9}{4(\frac{r}{n}) +3}) )$
$ = \int_{0}^1  \frac{16x +9}{4x+3} dx$
$ = 4 + \frac{3}{4} ln \frac{3}{7} $
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