비교 판정법

(a)

• Solution 1: The given series is

$$\begin{aligned} 1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\cdots =\sum_{n=1}^\infty a_n \text{ with $a_n=\frac{1}{2n-1}$} \end{aligned}$$

First we'll develop some intuition by observing that, for very large $n$, $a_n\approx \frac{1}{2n}$. We know that the series $\sum\limits_{n=1}^\infty\frac{1}{n}$ diverges by the $p$–test with $p=1$. So let's apply the limit comparison test with $b_n=\frac{1}{n}$. Since

$$\begin{aligned} \lim_{n\rightarrow\infty}\frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\frac{n}{2n-1} =\lim_{n\rightarrow\infty}\frac{1}{2-\frac{1}{n}} =\frac{1}{2} \end{aligned}$$

the series $\sum\limits_{n=1}^\infty a_n$ converges if and only if the series $\sum\limits_{n=1}^\infty b_n$ converges. So the given series diverges.

• Solution 2: The series

$$\begin{aligned} 1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\cdots &\ge \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\cdots \\ &=\frac{1}{2}\Big(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots\Big) \end{aligned}$$

The series in the brackets is the harmonic series which we know diverges, by the $p$–test with $p=1$. So the series on the right hand side diverges. By the direct comparison test, the series on the left hand side diverges too.

(b) We'll use the ratio test with $a_n=\dfrac{(2n+1)}{2^{2n+1}}$. Since

$$\begin{aligned} \frac{a_{n+1}}{a_n} &=\frac{(2n+3)}{2^{2n+3}}\frac{2^{2n+1}}{(2n+1)} =\frac{1}{4}\frac{(2n+3)}{(2n+1)} =\frac{1}{4}\frac{(2+3/n)}{(2+1/n)} \rightarrow \frac{1}{4}<1\text{ as }n\rightarrow\infty \end{aligned}$$

the series converges.