Question

Let f be defined by the function f(x) = 1/(x^2+9)

(a) Evaluate the improper integral
`\int\limits^(∞)_3 {f(x)} \, dx` 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)

Answer 1

(a)

`\displaystyle\int_3^\infty (\mathrm dx)/(x^2+9)=\lim_(b\to\infty)\int_(x=3)^(x=b)(\mathrm dx)/(x^2+9)`

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

`\displaystyle\lim_(b\to\infty)\int_(t=\arctan(1))^(t=\arctan\left(\frac b3\right))(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`

`=\displaystyle \frac13 \lim_(b\to\infty)\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\frac\pi{12}}`

(b) The series

`\displaystyle \sum_(n=3)^\infty \frac1{n^2+9}`

converges by comparison to the convergent p-series,

`\displaystyle\sum_(n=3)^\infty\frac1{n^2}`

(c) The series

`\displaystyle \sum_(n=1)^\infty ((-1)^n (n^2+9))/(e^n)`

converges absolutely, since

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

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.