Calculus 8th Edition

sum of $\Sigma a_{n}$ (where n is an integer starting at 1 and approaching infinity) is: $\lim\limits_{n\to \infty} s_{n}$ = $\lim\limits_{n\to \infty} \frac{(n^2) -1 }{4(n^4) + 1}$ = $\lim\limits_{n\to \infty} \frac{n^2} {4(n^2) } \frac{1 -\frac{1 }{n^2} } {1 + \frac{1 }{n^2} }$= $\frac{1}{4}$
Recall the Definition 2 from page 748 in our textbook, taking the limit we will get: $\lim\limits_{n\to \infty} s_{n}$ = $\lim\limits_{n\to \infty} \frac{(n^2) -1 }{4(n^2) + 1}$ = $\frac{1}{4}$. Now, how we arrived at $\frac{1}{4}$as an answer only requires that we take the highest order of terms as common factor from denominator and numerator and then take the limit. We get $\frac{n^2} {4(n^2) } \frac{1 -\frac{1 }{n^2} } {1 + \frac{1 }{n^2} }$ ; then $\frac{n^2} {4(n^2) }$ equals $\frac{1} {4 }$ , taking the limit of this fraction for n approaching infinity gives us $\frac{1}{4}$and thus the whole summation is $\frac{1}{4}$.