Solutions to Problem Set 22

Math 211-03

10-30-2017

[Series review]

1. Determine whether the series $\displaystyle \sum_{n=1}^\infty \left(\dfrac{1}{n} -
   \dfrac{2}{2 n + 1}\right)$ converges or diverges.

Add the fractions over a common denominator:

$$\dfrac{1}{n} - \dfrac{2}{2 n + 1} = \dfrac{2 n + 1}{n(2 n + 1)} - \dfrac{2 n}{n(2 n + 1)} = \dfrac{1}{n(2 n + 1)}.$$

Therefore, the series is $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n(2 n + 1)}$ . Apply Limit Comparison:

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

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


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

Apply the Root Test:

$$\lim_{n \to \infty} \left[\left(\dfrac{3 n + 1}{4 n + 5}\right)^n\right]^{1/n} = \lim_{n \to \infty} \dfrac{3 n + 1}{4 n + 5} = \dfrac{3}{4}.$$

The limiting ratio is less than 1, so the series converges, by the Root Test.


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

The terms are positive. Let $f(x) = \dfrac{\sqrt{\tan^{-1} x}}{x^2 + 1}$ . Then f is continuous for $x \ge 1$ .

I have

$$f'(x) = \dfrac{1 - 4 x \tan^{-1} x}{2(x^2 + 1)^2 \sqrt{\tan^{-1} x}}.$$

Note that $\tan^{-1} 1 =
   \dfrac{\pi}{4}$ , so $1 - 4 x \tan^{-1} x = 1 - \pi < 0$ when $x = 1$ . But x and $\tan^{-1} x$ both increase for $x \ge 1$ . Therefore, $f'(x) < 0$ for $x \ge 1$ , so the terms of the series decrease.

The conditions for the Integral Test are satisfied. Compute the integral:

$$\int_1^\infty \dfrac{\sqrt{\tan^{-1} x}}{x^2 + 1}\,dx = \lim_{b \to \infty} \int_1^b \dfrac{\sqrt{\tan^{-1} x}}{x^2 + 1}\,dx = \lim_{b \to \infty} \int_{\pi/4}^{\tan^{-1} b} \dfrac{\sqrt{u}}{x^2 + 1} \cdot (x^2 + 1)\,du = \lim_{b \to \infty} \int_{\pi/4}^{\tan^{-1} b} \sqrt{u}\,du =$$

$$\left[u = \tan^{-1} x, \quad du = \dfrac{dx}{x^2 + 1}, \quad dx = (x^2 + 1)\,du; \quad x = 1, u = \dfrac{\pi}{4}; \quad x = b, u = \tan^{-1} b\right]$$

$$\lim_{b \to \infty} \left[\dfrac{2}{3} u^{3/2}\right]_{\pi/4}^{\tan^{-1} b} = \dfrac{2}{3} \lim_{b \to \infty} \left((\tan^{-1} b)^{3/2} - \left(\dfrac{\pi}{4}\right)^{3/2}\right) = \dfrac{2}{3} \left(\left(\dfrac{\pi}{2}\right)^{3/2} - \left(\dfrac{\pi}{4}\right)^{3/2}\right).$$

Since the integral converges, the series converges by the Integral Test.


4. Determine whether the following series converges or diverges:

$$\dfrac{1}{1 \cdot \root 3 \of 1} + \dfrac{1}{2 \cdot \root 3 \of 2} + \dfrac{1}{3 \cdot \root 3 \of 3} + \dfrac{1}{4 \cdot \root 3 \of 4} + \cdots.$$

The series is $\displaystyle
   \sum_{n=1}^\infty \dfrac{1}{n^{4/3}}$ . It is a p-series with $p = \dfrac{4}{3} > 1$ , so it converges.


5. Determine whether the following series converges or diverges:

$$\sin \dfrac{\pi}{3} + \sin \dfrac{\pi}{6} + \sin \dfrac{\pi}{9} + \sin \dfrac{\pi}{12} + \cdots.$$

Write the series as $\displaystyle \sum_{n=1}^\infty \sin \dfrac{\pi}{3 n}$ . I'll use Limit Comparison:

$$\lim_{n \to \infty} \dfrac{\sin \dfrac{\pi}{3 n}}{\dfrac{\pi}{3 n}} = \lim_{u\to 0} \dfrac{\sin u}{u} = 1.$$

$$\left[u = \dfrac{\pi}{3 n}; \quad\hbox{as}\quad n \to \infty, \quad u\to 0\right]$$

$\displaystyle \sum_{n=1}^\infty
   \dfrac{\pi}{3 n}$ is $\dfrac{\pi}{3}$ times the harmonic series, so it diverges. Therefore, the original series diverges by Limit Comparison.


6. Determine whether the series $\displaystyle \sum_{n=1}^\infty \dfrac{5^n}{(n + 1)4^n}$ converges or diverges.

Use the Ratio Test:

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

The limiting ratio is greater than 1, so the series diverges.


7. Determine whether the series $\displaystyle \sum_{n=1}^\infty \dfrac{2^n + 4^n}{3^n + 5 \cdot
   4^n}$ converges or diverges.

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

($\left\{\dfrac{2^n}{4^n}\right\}$ and $\left\{\dfrac{3^n}{4^n}\right\}$ are geometric sequences with ratios $\dfrac{2}{4}$ and $\dfrac{3}{4}$ . Since the ratios are less than 1, the sequences go to 0 as n goes to $\infty$ .)

Since $\dfrac{2^n + 4^n}{3^n + 5
   \cdot 4^n}$ does not approach 0 as n goes to $\infty$ , the series diverges by the Zero Limit Test.


8. Use direct comparison to show that the following series converges: $\displaystyle \sum_{n=2}^\infty
   \dfrac{n - 1}{n^3 + 2}$ .

Changing the top from $n - 1$ to $(n - 1) + 1 = n$ makes the top bigger, so the fraction gets bigger. Changing $n^3 + 2$ to $(n^3 + 2) -
   2 = n^3$ makes the bottom smaller, so the fraction gets bigger. Together, these changes make the fraction bigger:

$$\dfrac{n - 1}{n^3 + 2} < \dfrac{n}{n^3} = \dfrac{1}{n^2}.$$

The series $\displaystyle
   \sum_{n=1}^\infty \dfrac{1}{n^2}$ converges, because it's a p-series with $p = 2 > 1$ . Therefore, the original series converges by direct comparison.


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