Solutions - 0180-04

(a) Verify applicability of the integral test:

• $f(x)$ is continuous for all $x\ge 1$. (Only discontinuity is at $x=0$, but the series starts at $x=1$.)
• $f(x)\ge 0$ since $x^{1.1}\ge 0$ for all $x\ge 1$.
• $f(x)$ is monotone decreasing, since as $x$ increases, the denominator increases, and the term decreases.

Apply the integral test:

$$\begin{array}{c}{\int }_1^{\infty }\frac{dx}{x^{1.1}}=\underset{R\to \infty }{\lim }{\int }_1^R\frac{dx}{x^{1.1}}\\ \\ \gg \gg \underset{R\to \infty }{\lim }[-10x^{-0.1}]{|}_1^R\gg \gg 0--10=10\end{array}$$

This is finite and the improper integral converges, so the series converges by the Integral Test.

---

(b) Verify applicability of the integral test with $f(x)=x{e}^{-x^2}$:

• $f(x)$ is definitely continuous for all $x\in ℝ$.
• $f(x)>0$ since $x>1$ and $e^{-x^2}>0$ for all $x\in ℝ$.
• Decreasing?
  • $f^{\prime }(x)=e^{-x^2}-2x^2{e}^{-x^2}=(1-2x^2)e^{-x^2}$ has zeros at $x=\pm \frac{1}{\sqrt{2}}$.
  • When $x>\frac{1}{\sqrt{2}}$, $f^{\prime }(x)<0$.
  • Series starts at $n=1>\frac{1}{\sqrt{2}}$, so the terms are monotone decreasing.

Apply the integral test:

$$\begin{array}{c}{\int }_1^{\infty }x{e}^{-x^2}dx=\underset{R\to \infty }{\lim }{\int }_1^{R}x{e}^{-x^2}dx\\ \\ \gg \gg \underset{R\to \infty }{\lim }{[-\frac{e^{-x^2}}{2}]|}_1^R\gg \gg 0--\frac{e^{-1}}{2}\gg \gg \frac{1}{2e}\end{array}$$

This is finite and the improper integral converges, so the series converges by the Integral Test.

---

(c) Verify applicability of the integral test for $f(x)=x^{-2/3}$:

• $f(x)$ is continuous for all $x\ge 1$. (The only discontinuity is at $x=0$, but the series starts at $n=1$.)
• $f(x)>0$ since $x^{-2/3}>0$ for all $x>1$.
• $f(x)$ is monotone decreasing, since as $x$ increases, the denominator increases, and the term decreases.

Apply the integral test:

$$\begin{array}{c}{\int }_1^{\infty }\frac{dx}{x^{2/3}}=\underset{R\to \infty }{\lim }{\int }_1^R\frac{dx}{x^{2/3}}\\ \\ \gg \gg \underset{R\to \infty }{\lim }\left({3x^{1/3}|}_1^R\right)\gg \gg \infty -3\end{array}$$

So the improper integral diverges, and the series diverges by the Integral Test.