비교 판정법

(a) For large $n$, $n^2\gg n+1$ and so the numerator $n^2+n+1\approx n^2$. For large $n$, $n^5\gg n$ and so the denominator $n^5-n\approx n^5$. So, for large $n$, $$\frac{n^2+n+1}{n^5-n}\approx \frac{n^2}{n^5} =\frac{1}{n^3}.$$ This suggests that we apply the limit comparison test with $a_n=\frac{n^2+n+1}{n^5-n}$ and $b_n=\frac{1}{n^3}$. Since

$$\begin{aligned} \lim_{n\rightarrow\infty} \frac{a_n}{b_n} &=\lim_{n\rightarrow\infty} \frac{(n^2+n+1)/(n^5-n)}{1/n^3} =\lim_{n\rightarrow\infty} \frac{n^5+n^4+n^3}{n^5-n} =\lim_{n\rightarrow\infty} \frac{1+1/n+1/n^2}{1-1/n^4} \\ &=1 \end{aligned}$$

exists and is nonzero, and since $\sum\limits_{n=1}^\infty\frac{1}{n^3}$ converges (by the $p$–test with $p=3>1$), the given series converges.

(b) For large $m$, $3m\gg |\sin\sqrt{m}|$ and so $$\frac{3m+\sin\sqrt{m}}{m^2}\approx \frac{3m}{m^2} =\frac{3}{m}.$$ This suggests that we apply the limit comparison test with $a_m=\frac{3m+\sin\sqrt{m}}{m^2}$ and $b_m=\frac{1}{m}$. (We could also use $b_m=\frac{3}{m}$.) Since

$$\begin{aligned} \lim_{m\rightarrow\infty} \frac{a_m}{b_m} &=\lim_{m\rightarrow\infty} \frac{(3m+\sin\sqrt{m})/m^2}{1/m} =\lim_{m\rightarrow\infty} \frac{3m+\sin\sqrt{m}}{m} =\lim_{m\rightarrow\infty} 3+\frac{\sin\sqrt{m}}{m} \\ &=3 \end{aligned}$$

exists and is nonzero, and since $\sum\limits_{m=1}^\infty\frac{1}{m}$ diverges (by the $p$–test with $p=1$), the given series diverges.