Science:Math Exam Resources/Courses/MATH105/April 2015/Question 05 (b)/Solution 1

We will apply the comparison test to show that the series converges. The comparison test states that if $a_{n}$ and $b_{n}$ are two sequences with $0\leq b_{n}\leq a_{n}$ for $n\geq k$, then if $\sum _{n=k}^{\infty }a_{n}$ converges, the series $\sum _{n=k}^{\infty }b_{n}$ converges as well.

Since $n\geq 2$, $1\leq n^{2}$ and $n\leq n^{2}$. Hence $n^{2}+n+1\leq n^{2}+n^{2}+n^{2}=3n^{2}$.

Furthermore, since $n\leq {\frac {1}{2}}n^{5}$ for all $n\geq 2$, we have $n^{5}-n\geq {\frac {1}{2}}n^{5}$.

Therefore,

${\frac {n^{2}+n+1}{n^{5}-n}}\leq {\frac {3n^{2}}{{\frac {1}{2}}n^{5}}}={\frac {6}{n^{3}}}.$

Since $\sum _{n=2}^{\infty }{\frac {6}{n^{3}}}$ converges by the p-series test (here, $p=3$), by comparison the series $\sum _{n=2}^{\infty }{\frac {n^{2}+n+1}{n^{5}-n}}$ also converges.