비교 판정법

• Solution 1: Let's see whether the divergence test works here. $$\lim_{n\to\infty} \frac{n^4 2^{n/3}}{(2n+7)^4}\left(\frac{\frac{1}{n^4}}{\frac{1}{n^4}}\right) = \lim_{n\to\infty} \frac{2^{n/3}}{(2+7/n)^4} = \lim_{n\to\infty} \frac{2^{n/3}}{(2+0)^4} = \infty$$ The summands of our series do not converge to zero. By the divergence test, the series diverges.
• Solution 2: Let's develop some intuition for a comparison. For very large $n$, $2n$ dominates $7$ so that

$$\begin{aligned} \frac{n^4 2^{n/3}}{(2n+7)^4} \approx \frac{n^4 2^{n/3}}{(2n)^4} =\frac{1}{16}2^{n/3} \end{aligned}$$

The series $\displaystyle \sum_{n=1}^{\infty} 2^{n/3}$ is a geometric series with ratio $r=2^{1/3}>1$ and so diverges. (It also fails the divergence test.) We expect the given series to diverge too.

To verify that our intuition is correct, we apply the limit comparison test with

$$\begin{aligned} a_n= \frac{n^4 2^{n/3}}{(2n+7)^4} \quad\text{and}\quad b_n= 2^{n/3} \end{aligned}$$

which is valid since

$$\lim_{n\rightarrow\infty} \frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\frac{n^4}{(2n+7)^4} =\lim_{n\rightarrow\infty}\frac{1}{{(2+7/n)}^4} =\frac{1}{2^4}$$

exists and is nonzero. Since the series $\sum\limits_{n=1}^\infty b_n$ is a divergent geometric series (with ratio $r=2^{1/3}>1$), the given series diverges.

(It is possible to use the plain comparison test as well. One needs to show something like $a_n = \frac{n^4 2^{n/3}}{(2n+7)^4} \ge \frac{n^4 2^{n/3}}{(2n+7n)^4} = \frac{1}{9^4}b_n$.)

• Solution 3: Alternately, one can apply the ratio test:

$$\begin{aligned} \lim_{n\to\infty} \bigg| \frac{a_{n+1}}{a_n} \bigg| &= \lim_{n\to\infty} \bigg| \frac{(n+1)^42^{(n+1)/3}/(2(n+1)+7)^4}{n^42^{n/3}/(2n+7)^4} \bigg| \\ &= \lim_{n\to\infty} \frac{(n+1)^4(2n+7)^4}{n^4(2n+9)^4}\ \frac{2^{(n+1)/3}}{2^{n/3}} \\ &=\lim_{n\to\infty} \frac{(1+1/n)^4(2+7/n)^4}{(2+9/n)^4} \cdot2^{1/3} = 1\cdot2^{1/3} > 1. \end{aligned}$$

Since the ratio of consecutive terms is greater than one, by the ratio test, the series diverges.