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

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

Since $\sin {\sqrt {m}}\geq -1$ for all $m\geq 1$, $3m+\sin {\sqrt {m}}\geq 3m-1$. Furthermore, $3m-1\geq 2m$ for all $m\geq 1$, so

${\frac {3m+\sin {\sqrt {m}}}{m^{2}}}\geq {\frac {3m-1}{m^{2}}}\geq {\frac {2m}{m^{2}}}={\frac {2}{m}}.$

As $\sum _{m=1}^{\infty }{\frac {2}{m}}$ diverges by the p-series test (here, $p=1$), then by the comparison test, the series $\sum _{m=1}^{\infty }{\frac {3m+\sin {\sqrt {m}}}{m^{2}}}$ diverges as well.