test the series for convergence or divergence. ∑(n = 2 to ∞) 1 / (n√ln(8n)) convergent divergent need help…

Explanation:

Step1: Apply integral - test

The integral - test states that for a series $\sum_{n = N}^{\infty}a_{n}$ where $a_{n}=f(n)$, $f(x)$ is positive, continuous and decreasing for $x\geq N$, the series $\sum_{n = N}^{\infty}a_{n}$ and the improper integral $\int_{N}^{\infty}f(x)dx$ either both converge or both diverge. Let $f(x)=\frac{1}{x\sqrt{\ln(8x)}}$, which is positive, continuous and decreasing for $x\geq2$.

Step2: Evaluate the improper integral

Let $u = \ln(8x)$, then $du=\frac{1}{x}dx$. When $x = 2$, $u=\ln(16)$ and as $x\rightarrow\infty$, $u\rightarrow\infty$. The integral $\int_{2}^{\infty}\frac{1}{x\sqrt{\ln(8x)}}dx=\int_{\ln(16)}^{\infty}\frac{1}{\sqrt{u}}du$.

Step3: Calculate the integral of $\frac{1}{\sqrt{u}}$

We know that $\int\frac{1}{\sqrt{u}}du=\int u^{-\frac{1}{2}}du = 2u^{\frac{1}{2}}+C$. Then $\int_{\ln(16)}^{\infty}\frac{1}{\sqrt{u}}du=\lim_{t\rightarrow\infty}\int_{\ln(16)}^{t}u^{-\frac{1}{2}}du=\lim_{t\rightarrow\infty}(2\sqrt{u}\big|_{\ln(16)}^{t})$.

Step4: Find the limit

$\lim_{t\rightarrow\infty}(2\sqrt{u}\big|{\ln(16)}^{t})=\lim{t\rightarrow\infty}(2\sqrt{t}-2\sqrt{\ln(16)})=\infty$.

Answer:

divergent