Answered

Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

Let f be defined by the function f(x) = 1/(x^2+9)
(a) Evaluate the improper integral [tex]\int\limits^{∞}_3 {f(x)} \, dx[/tex] or show that the integral diverges
(b) Determine whether the series ∑n=3∞ f(n) converges or diverges State the conditions of the test used for determining convergence or divergence
(c) Determine whether the series ∑n=1∞(−1)n(en⋅f(n))=∑n=1∞(−1)n(n2+9)en converges absolutely, converges conditionally, or diverges (image put below)

Let F Be Defined By The Function Fx 1x29 A Evaluate The Improper Integral Texintlimits3 Fx Dxtex Or Show That The Integral Diverges B Determine Whether The Seri class=

Sagot :

LammettHash

(a)

[tex]\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}[/tex]

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

[tex]\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt[/tex]

[tex]=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}[/tex]

(b) The series

[tex]\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}[/tex]

converges by comparison to the convergent p-series,

[tex]\displaystyle\sum_{n=3}^\infty\frac1{n^2}[/tex]

(c) The series

[tex]\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}[/tex]

converges absolutely, since

[tex]\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}[/tex]

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.