Solutions to Problem Set 21

Math 211-03

10-30-2017

[Ratio and root test]

1. Use the Ratio Test to determine whether the series $\displaystyle \sum_{n=1}^\infty
   \dfrac{2^n + 3^n}{4^n + 5^n}$ converges or diverges.

$$\lim_{n \to \infty} \dfrac{a_{n+1}}{a_n} = \lim_{n \to \infty} \dfrac{\dfrac{2^{n+1} + 3^{n+1}}{4^{n+1} + 5^{n+1}}}{\dfrac{2^n + 3^n}{4^n + 5^n}} = \lim_{n \to \infty} \dfrac{2^{n+1} + 3^{n+1}}{4^{n+1} + 5^{n+1}} \cdot \dfrac{4^n + 5^n}{2^n + 3^n} = \lim_{n \to \infty} \dfrac{2^{n+1} + 3^{n+1}}{2^n + 3^n} \cdot \dfrac{4^n + 5^n}{4^{n+1} + 5^{n+1}} =$$

$$\lim_{n \to \infty} \dfrac{2^{n+1} + 3^{n+1}}{2^n + 3^n} \cdot \dfrac{\dfrac{1}{3^{n+1}}}{\dfrac{1}{3^{n+1}}} \cdot \dfrac{4^n + 5^n}{4^{n+1} + 5^{n+1}} \cdot \dfrac{\dfrac{1}{5^{n+1}}}{\dfrac{1}{5^{n+1}}} = \lim_{n \to \infty} \dfrac{\dfrac{2^{n+1}}{3^{n+1}} + 1}{\dfrac{1}{3} \cdot \dfrac{2^n}{3^n} + \dfrac{1}{3}} \cdot \dfrac{\dfrac{1}{5} \cdot \dfrac{4^n}{5^n} + \dfrac{1}{5}}{\dfrac{4^{n+1}}{5^{n+1}} + 1} = \dfrac{0 + 1}{0 + \dfrac{1}{3}} \cdot \dfrac{0 + \dfrac{1}{5}} {0 + 1} = \dfrac{\dfrac{1}{5}}{\dfrac{1}{3}} = \dfrac{3}{5} < 1.$$

In computing the limit, I used the facts that

$$\lim_{n \to \infty} \dfrac{2^{n+1}}{3^{n+1}} = 0, \quad \lim_{n \to \infty} \dfrac{2^n}{3^n} = 0, \quad \lim_{n \to \infty} \dfrac{4^n}{5^n} = 0, \quad\hbox{and}\quad \lim_{n \to \infty} \dfrac{4^{n+1}}{5^{n+1}} = 0.$$

The reason is that all four are geometric sequences with ratios between -1 and 1.

The series converges by the Ratio Test.


2. Use the Ratio Test to determine whether the series $\displaystyle \sum_{n=1}^\infty
   \dfrac{5^n (n!)^2}{(2 n)!}$ converges or diverges.

$$\lim_{n \to \infty} \dfrac{a_{n+1}}{a_n} = \lim_{n \to \infty} \dfrac{\dfrac{5^{n+1} ((n + 1)!)^2}{(2 n + 2)!}} {\dfrac{5^n (n!)^2}{(2 n)!}} = \lim_{n \to \infty} \dfrac{5^{n+1} ((n + 1)!)^2}{(2 n + 2)!} \cdot \dfrac{(2 n)!}{5^n (n!)^2} = \lim_{n \to \infty} \dfrac{5^{n+1}}{5^n} \cdot \dfrac{((n + 1)!)^2}{(n!)^2} \cdot \dfrac{(2 n)!}{(2 n + 2)!} =$$

$$\lim_{n \to \infty} 5 \cdot \left(\dfrac{1 \cdot 2 \cdots n \cdot (n + 1)}{ 1\cdot 2 \cdots n}\right)^2 \cdot \dfrac{1 \cdot 2 \cdots (2 n)}{1 \cdot 2 \cdots (2 n) \cdot (2 n + 1) \cdot (2 n + 2)} = \lim_{n \to \infty} 5 \cdot (n + 1)^2 \cdot \dfrac{1}{(2 n + 1)(2 n + 2)} =$$

$$\lim_{n \to \infty} \dfrac{5 (n + 1)^2}{(2 n + 1)(2 n + 2)} = \dfrac{5}{4} > 1.$$

The series diverges by the Ratio Test.


3. (a) Show that the Ratio Test fails when it is applied to the series $\displaystyle
   \sum_{n=1}^\infty \dfrac{n + 2}{n^3 + 3 n + 1}$ .

(b) Use Limit Comparison to determine whether the series $\displaystyle \sum_{n=1}^\infty
   \dfrac{n + 2}{n^3 + 3 n + 1}$ converges or diverges.

(a)

$$\lim_{n \to \infty} \dfrac{a_{n+1}}{a_n} = \lim_{n \to \infty} \dfrac{\dfrac{n + 3}{(n + 1)^3 + 3 (n + 1) + 1}}{\dfrac{n + 2}{n^3 + 3 n + 1}} = \lim_{n \to \infty} \dfrac{n + 3}{(n + 1)^3 + 3 (n + 1) + 1} \cdot \dfrac{n^3 + 3 n + 1}{n + 2} =$$

$$\lim_{n \to \infty} \dfrac{n + 3}{n + 2} \cdot \dfrac{n^3 + 3 n + 1}{(n + 1)^3 + 3 (n + 1) + 1} = 1 \cdot 1 = 1.$$

Since the limit is 1, the Ratio Test fails.

Note: The point here is that if the general term of the series is "just powers", you should not use the Ratio Test.

(b)

$$\lim_{n \to \infty} \dfrac{\dfrac{n + 2}{n^3 + 3 n + 1}}{\dfrac{1}{n^2}} = \lim_{n \to \infty} \dfrac{n + 2}{n^3 + 3 n + 1} \cdot \dfrac{n^2}{1} = \lim_{n \to \infty} \dfrac{n^3 + 2 n^2}{n^3 + 3 n + 1} = 1.$$

The limit is a finite positive number. $\displaystyle \sum_{n=1}^\infty
   \dfrac{1}{n^2}$ converges, because it's a p-series with $p
   = 2 > 1$ . Therefore, the original series diverges by Limit Comparison.


4. Use the Root Test to determine whether the series $\displaystyle \sum_{n=1}^\infty
   \left(\dfrac{3 n + 1}{7 n + 2}\right)^n$ converges or diverges.

$$\lim_{n \to \infty} \left(\left(\dfrac{3 n + 1}{7 n + 2}\right)^n\right)^{1/n} = \lim_{n \to \infty} \dfrac{3 n + 1}{7 n + 2} = \dfrac{3}{7} < 1.$$

The series converges by the Root Test.


5. Use the Root Test to determine whether the series $\displaystyle \sum_{n=1}^\infty
   \left(1 - \dfrac{3}{n}\right)^{n^2}$ converges or diverges.

$$\lim_{n \to \infty} \left(\left(1 - \dfrac{3}{n}\right)^{n^2}\right)^{1/n} = \lim_{n \to \infty} \left(1 - \dfrac{3}{n}\right)^{n}.$$

Let $y = \left(1 -
   \dfrac{3}{n}\right)^{n}$ . Then

$$\ln y = \ln \left(1 - \dfrac{3}{n}\right)^{n} = n \ln \left(1 - \dfrac{3}{n}\right).$$

So

$$\lim_{n \to \infty} \ln y = \lim_{n \to \infty} n \ln \left(1 - \dfrac{3}{n}\right) = \lim_{n \to \infty} \dfrac{\ln \left(1 - \dfrac{3}{n}\right)} {\dfrac{1}{n}} = \lim_{n \to \infty} \dfrac{\left(\dfrac{1}{1 - \dfrac{3}{n}}\right) \left(\dfrac{3}{n^2}\right)}{-\dfrac{1}{n^2}} = -3 \lim_{n \to \infty} \dfrac{1}{1 - \dfrac{3}{n}} = -3.$$

Therefore,

$$\lim_{n \to \infty} \left(1 - \dfrac{3}{n}\right)^{n} = e^{-3} < 1.$$

The series converges by the Root Test.


6. Use the Root Test to determine whether the series $\displaystyle \sum_{n=1}^\infty
   \dfrac{1}{(\ln n)^n}$ converges or diverges.

$$\lim_{n \to \infty} \left(\dfrac{1}{(\ln n)^n}\right)^{1/n} = \lim_{n \to \infty} \dfrac{1}{\ln n} = 0 < 1.$$

The series converges by the Root Test.


The trouble with the rat race is that even if you win you're still a rat. - Lily Tomlin


Contact information

Bruce Ikenaga's Home Page

Copyright 2017 by Bruce Ikenaga