Question

2 points) Test each of the following series for convergence by the Integral Test. If the Integral Test can be applied to the series, enter CONV if it converges or DIV if it diverges. If the integral test cannot be applied to the series, enter NA. (Note: this means that even if you know a given series converges by some other test, but the Integral Test cannot be applied to it, then you must enter NA rather than CONV.) CONV 1. ∑n=1[infinity]9n(ln(9n))5 DIV 2. ∑n=1[infinity]ne−8n NA 3. ∑n=1[infinity]ne8n DIV 4. ∑n=1[infinity]ln(3n)n NA 5. ∑n=1[infinity]n+7(−4)n

Answer 1

Answer:

Required result : (a) Divergent. (b) Convergent. (c) Divergent, (d) Not exists. (e) Not exists.

Step-by-step explanation:

By the definition of integral test the series for a integer n and a continuous function f(x) which is monotonic decreasing in $[n,\infty)$ then the infinite series $\sum_{n}^{\infty}$ converges or diverges if and only if the improper integral $\int_{n}^{\infty}$ converges or diverges.

Given,

(a) $\sum_{n=1}^{\infty}9^n(\ln(9^n))^n$

$=5(\ln(9))^5\sum_{n=1}^{\infty}9^nn^5\hfill (1)$

Then,

$\int_{1}^{\infty}9^n(\ln(9^n))^5$

$=5(\ln 9)^5\int_{1}^{\infty}9^nn^5dn\hfill (2)$

Let,

$I=\int 9^nn^9$

By using integral calculator we get,

$I=\dfrac{\left(2\ln\left(3\right)n\left(\ln\left(3\right)n\left(\ln\left(3\right)n\left(\ln\left(3\right)n\left(2\ln\left(3\right)n-5\right)+10\right)-15\right)+15\right)-15\right)\mathrm{e}^{2\ln\left(3\right)n}}{8\ln^6\left(3\right)}$

which is divergent. Therefore given (2) is divergent and so is (1).

(b) $\sum_{n=1}^{\infty}ne^{-8n}\hfill (3)$

Then in integral form,

$\int_{1}^{\infty}ne^{-8n}$

$=\frac{9e^{-8}}{64}$

$=4.72\times 10^{-5}$

Thus given series (3) is convergent.

(c) $\sum_{n=1}^{\infty}ne^{8n}\hfill (4)$

Then in integral form,

$\int_{1}^{\infty}ne^{-8n}$

Now let,

$I=\int n e^{-8n}$

Applying integral calculator we get,

$I=\dfrac{\left(8n-1\right)\mathrm{e}^{8n}}{64}$

which is divergent and thus,

$\int_{1}^{\infty}I$ is divergent. So, given series (4) is divergent.

(d) $\sun_{n=1}^{\infty}ln(3n)^n$

which is in integral form,

$\int_{1}^{\infty}\ln (3n)^n$

during integration since there is no any antiderivative, the result could not be found.

(e)  $\sum_{n=1}^{\infty}n+7(-4)^n$

Integral form is,

$\int_{1}^{\infty}n+7^{(-4)^n}$

Let,

$I=\int _{1}^{b}n+7^{(-4)^n}$

Using integral calculator we get,

$I=\frac{1}{2\ln(-4)}\Big[\ln(-4)b^2-2\Gamma(0,-\ln(7)(-4)^b)+2\Gamma(0,4\ln(7))-\ln(-4)\Big]$

But,

$\lim_{b\to \infty}I$ not exists.