Science:Math Exam Resources/Courses/MATH101/April 2012/Question 04 (a)/Solution 1

We will compare our series with $\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{4}}}.$ So we look at the limit:

${\begin{aligned}\lim _{n\to \infty }{\frac {({\frac {n^{2}-\sin(n)}{n^{6}+n^{2}}})}{({\frac {1}{n^{4}}})}}&=\lim _{n\to \infty }{\frac {n^{6}+n^{4}\sin(n)}{n^{6}+n^{2}}}\\&=\lim _{n\to \infty }{\frac {1-{\frac {\sin(n)}{n^{2}}}}{1+{\frac {1}{n^{4}}}}}\\&={\frac {1}{1}}=1\end{aligned}}$

Since $\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{4}}}$ converges, by limit comparison our series also converges.