Solutions to Problem Set 18

Math 211-03

10-23-2017

[Integral Test]

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

The series has positive terms.

Let $f(x) = \dfrac{1}{4 x + 7}$ . Then f is continuous for $x \ge 1$ .

Now

$$f'(x) = -\dfrac{4}{(4 x + 7)^2}.$$

Since $f'(x) < 0$ for $x
   \ge 1$ , the terms of the series decrease.

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

$$\int_1^\infty \dfrac{1}{4 x + 7}\,dx = \lim_{a \to \infty} \int_1^a \dfrac{1}{4 x + 7}\,dx = \lim_{a \to \infty} \left[\dfrac{1}{4} \ln |4 x + 7|\right]_1^a =$$

$$\dfrac{1}{4} \lim_{a \to \infty} \left(\ln |4 a + 7| - \ln 11\right) = +\infty.$$

The integral diverges, so the series diverges, by the Integral Test.


2. Determine whether the series $\displaystyle \sum_{n=2}^\infty \dfrac{e^{1/n}}{n^2}$ converges or diverges.

The series has positive terms.

Let $f(x) =
   \dfrac{e^{1/x}}{x^2}$ . Then f is continuous for $x \ge 2$ .

Now

$$f'(x) = -\dfrac{e^{1/x}}{x^4} - \dfrac{2 e^{1/x}}{x^3}.$$

Since $f'(x) < 0$ for $x
   \ge 2$ , the terms of the series decrease.

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

$$\int_2^\infty \dfrac{e^{1/x}}{x^2}\,dx = \lim_{a \to \infty} \int_{1/2}^a \dfrac{e^{1/x}}{x^2}\,dx = \lim_{a \to \infty} \int_{1/2}^{1/a} \dfrac{e^u}{x^2}\cdot (-x^2\,du) = -\lim_{a \to \infty} \int_{1/2}^{1/a} e^u\,du =$$

$$\left[u = \dfrac{1}{x}, \quad du = -\dfrac{1}{x^2}\,dx, \quad dx = -x^2\,du; \quad x = 2, u = 1/2; \quad x = a, u = \dfrac{1}{a}\right]$$

$$-\lim_{a \to \infty} \left[e^u\right]_{1/2}^{1/a} = -\lim_{a \to \infty} \left(e^{1/a} - e^{1/2}\right) = -(1 - e^{1/2}) = e^{1/2} - 1.$$

The integral converges, so the series converges, by the Integral Test.


3. Determine convergence or divergence of the series $\displaystyle \sum_{n=3}^\infty
   \dfrac{\ln n}{n}$ .

The series has positive terms.

Let $f(x) = \dfrac{\ln x}{x}$ . Then f is continuous for $x \ge 3$ .

Now

$$f'(x) = \dfrac{1}{x^2} - \dfrac{\ln x}{x^2} = \dfrac{1 - \ln x}{x^2}.$$

$f'(x) = 0$ for $x = e
   \approx 2.71828$ , and $f'(x) < 0$ for $x > e$ . The series starts at $n = 3$ , so the terms of the series decrease.

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

$$\int_3^\infty \dfrac{\ln x}{x}\,dx = \lim_{a \to \infty} \int_3^a \dfrac{\ln x}{x}\,dx = \lim_{a \to \infty} \int_{\ln 3}^{\ln a} \dfrac{u}{x}\cdot x\,du = \lim_{a \to \infty} \int_{\ln 3}^{\ln a} u\,du =$$

$$\left[u = \ln x, \quad du = \dfrac{dx}{x}, \quad dx = x\,du; \quad x = 3, u = \ln 3; \quad x = a, u = \ln a\right]$$

$$\lim_{a \to \infty} \left[\dfrac{1}{2} u^2\right]_{\ln 3}^{\ln a} = \dfrac{1}{2} \lim_{a \to \infty} \left((\ln a)^2 - (\ln 3)^2\right) = +\infty.$$

The integral diverges, so the series diverges, by the Integral Test.


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

The series has positive terms.

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

Now

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

If $x \ge 1$ , then $\tan^{-1} x \ge \dfrac{\pi}{4}$ , so

$$2 x \tan^{-1} x \ge 2 \cdot \dfrac{\pi}{4} = \dfrac{\pi}{2}, \quad\hbox{and}\quad 1 - 2 x \tan^{-1} x \le 1 - \dfrac{\pi}{2} < 0.$$

Since the top of the fraction for $f'(x)$ is negative and the bottom is positive, $f'(x) < 0$ for $x \ge 1$ . Therefore, the terms of the series decrease.

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

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

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

$$\lim_{a \to \infty} \left[\dfrac{1}{2} u^2\right]_{\pi/4}^{\tan^{-1} a} = \dfrac{1}{2} \lim_{a \to \infty} \left(\left(\tan^{-1} a\right)^2 - \dfrac{\pi^2}{16}\right) = \dfrac{1}{2} \left(\dfrac{\pi^2}{4} - \dfrac{\pi^2}{16}\right) = \dfrac{3 \pi^2}{32}.$$

The integral converges, so the series converges, by the Integral Test.


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