Solutions to Problem Set 20

Math 211-03

10-26-2017

[Limit comparison]

1. Use Limit Comparison to determine whether the series $\displaystyle \sum_{n=1}^\infty
   \dfrac{3 n^2 + 5}{17 n^4 + 10}$ converges or diverges.

$$\lim_{n \to \infty} \dfrac{\dfrac{3 n^2 + 5}{17 n^4 + 10}} {\dfrac{1}{n^2}} = \lim_{n \to \infty} \dfrac{3 n^2 + 5}{17 n^4 + 10} \cdot \dfrac{n^2}{1} = \lim_{n \to \infty} \dfrac{3 n^4 + 5 n^2}{17 n^4 + 10} = \dfrac{3}{17}.$$

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 converges by Limit Comparison.


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

$$\lim_{n \to \infty} \dfrac{\dfrac{\sqrt{n} + 5}{3 n + 4}}{\dfrac{1}{\sqrt{n}}} = \lim_{n \to \infty} \dfrac{\sqrt{n} + 5}{3 n + 4} \cdot \dfrac{\sqrt{n}}{1} = \lim_{n \to \infty} \dfrac{n + 5 \sqrt{n}}{3 n + 4} = \dfrac{1}{3}.$$

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


3. Use Limit Comparison to determine whether the series $\displaystyle \sum_{n=1}^\infty
   \dfrac{4^n + 7^n}{8^n + 2^n}$ converges or diverges.

$$\lim_{n \to \infty} \dfrac{\dfrac{4^n + 7^n}{8^n + 2^n}}{\dfrac{7^n}{8^n}} = \lim_{n \to \infty} \dfrac{4^n + 7^n}{8^n + 2^n} \cdot \dfrac{8^n}{7^n} = \lim_{n \to \infty} \dfrac{32^n + 56^n}{56^n + 14^n} = 1.$$

The limit is a finite positive number. $\displaystyle \sum_{n=1}^\infty
   \dfrac{7^n}{8^n}$ converges, because it's a geometric series with ratio $\dfrac{7}{8}$ , and $-1 <
   \dfrac{7}{8} < 1$ . Therefore, the original series converges by Limit Comparison.


4. Determine whether the series $\displaystyle \sum_{n=1}^\infty \dfrac{3 n^2 + 7 n + 1}{4 n^2 +
   2 n + 1}$ converges or diverges.

$$\lim_{n \to \infty} \dfrac{3 n^2 + 7 n + 1}{4 n^2 + 2 n + 1} = \dfrac{3}{4} \ne 0.$$

The series diverges by the Zero Limit Test.


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

$$\lim_{n \to \infty} \dfrac{\dfrac{n(n + 1)}{(n + 2)(n + 3)(n + 4)}} {\dfrac{1}{n}} = \lim_{n \to \infty} \dfrac{n(n + 1)}{(n + 2)(n + 3)(n + 4)} \cdot \dfrac{n}{1} = \lim_{n \to \infty} \dfrac{n^2(n + 1)}{(n + 2)(n + 3)(n + 4)} =$$

$$\lim_{n \to \infty} \dfrac{n^3 + n^2}{n^3 + 9 n^2 + 26 n + 24} = 1.$$

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


[Ratio and root test]

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

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

$$\lim_{n \to \infty} 5 \cdot \dfrac{1 \cdot 2 \cdots n} {1 \cdot 2 \cdots n \cdot (n + 1)} = \lim_{n \to \infty} \dfrac{5}{n + 1} = 0 < 1.$$

The series converges by the Ratio Test.


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

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

$$\lim_{n \to \infty} 3 \cdot \dfrac{(n + 1)^2}{n^2} \cdot \dfrac{1 \cdot 2 \cdots (n + 2)}{1 \cdot 2 \cdots (n + 2) \cdot (n + 3)} = \lim_{n \to \infty} \dfrac{3 (n + 1)^2}{n^2 (n + 3)} = 0 < 1.$$

(The limit is zero because the highest power on the top is $n^2$ , while the highest power on the bottom is $n^3$ .) The series converges by the Ratio Test.


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