Back to the class
Section 9.4 #4: Use the direct comparison test to determine the convergence or divergence of the series: $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{3n^2+2}$.
Solution: For all $n=1,2,\ldots$, $3n^2+2 \gt n^2$. Hence $\dfrac{1}{3n^2+2} \lt \dfrac{1}{n^2}$. Therefore $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{3n^2+2} \lt \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2}$. Since $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2}$ converges, we must conclude that $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{3n^2+2}$ converges.
Section 9.4 #5: Use the direct comparison test to determine the convergence or divergence of the series: $\displaystyle\sum_{n=2}^{\infty} \dfrac{1}{\sqrt{n}-1}$.
Solution: For $n=2,3,\ldots$, $\sqrt{n}-1 \lt \sqrt{n}$, hence $\dfrac{1}{\sqrt{n}-1} \gt \dfrac{1}{\sqrt{n}}$. Therefore $\displaystyle\sum_{n=2}^{\infty} \dfrac{1}{\sqrt{n}-1} \gt \displaystyle\sum_{n=2}^{\infty} \dfrac{1}{\sqrt{n}}$. But we know that $\displaystyle\sum_{n=2}^{\infty} \dfrac{1}{\sqrt{n}}$ diverges because it is a $p$-series with $p=\frac{1}{2}$ (also the integral test would show it), and therefore $\displaystyle\sum_{n=2}^{\infty} \dfrac{1}{\sqrt{n}-1}$ diverges.
Section 9.4 #14: Use the limit comparison test to determine the convergence or divergence of the series: $\displaystyle\sum_{n=1}^{\infty} \dfrac{5}{4^n+1}$.
Solution: We will compare to the sequence $\dfrac{1}{4^n}$ because we know that $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{4^n}$ converges (it is geometric with ratio $\frac{1}{4}$). For the limit comparison test, compute
$$\displaystyle\lim_{n \rightarrow \infty} \dfrac{\frac{5}{4^n+1}}{\frac{1}{4^n}} = 5 \displaystyle\lim_{n \rightarrow \infty} \dfrac{4^n}{4^n+1} = 5.$$
Since $5$ is a positive finite number, the limit comparison tests allows us to conclude that $\displaystyle\sum_{n=1}^{\infty} \dfrac{5}{4^n+1}$ converges since $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{4^n}$ converges.
Section 9.4 #15: Use the limit comparison test to determine the convergence or divergence of the series: $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n^2+1}}$.
Solution: We will compare to $\dfrac{1}{n}$. Compute
$$\displaystyle\lim_{n \rightarrow \infty} \dfrac{\frac{1}{\sqrt{n^2+1}}}{\frac{1}{n}}=\displaystyle\lim_{n \rightarrow \infty} \dfrac{n}{\sqrt{n^2+1}} = 1.$$
But since $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n}$ diverges (it is the harmonic series), the limit comparison test allows us to conclude that $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n^2+1}}$ also diverges.
Section 9.4 #17: Use the limit comparison test to determine the convergence or divergence of the series: $\displaystyle\sum_{n=1}^{\infty} \dfrac{2n^2-1}{3n^5+2n+1}$.
Solution: We will compare to $\dfrac{1}{n^3}$. Compute
$$\displaystyle\lim_{n \rightarrow \infty} \dfrac{\frac{1}{n^3}}{\frac{2n^2-1}{3n^5+2n+1}}= \dfrac{3}{2}.$$
Since $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^3}$ converges, the limit comparison test allows us to conclude that $\displaystyle\sum_{n=1}^{\infty} \dfrac{2n^2-1}{3n^5+2n+1}$ also converges.
