question 4 of 13\ndetermine convergence or divergence of $sum_{n = 2}^{infty}\frac{10}{nln(n)-n}$ using any…

question 4 of 13\ndetermine convergence or divergence of $sum_{n = 2}^{infty}\frac{10}{nln(n)-n}$ using any method.\nthe series diverges.\nthe series converges.

question 4 of 13\ndetermine convergence or divergence of $sum_{n = 2}^{infty}\frac{10}{nln(n)-n}$ using any method.\nthe series diverges.\nthe series converges.

Answer

Explanation:

Step1: Apply the integral test

We consider the function $f(x)=\frac{10}{x\ln(x) - x}$ for $x\geq2$. First, we can factor out $x$ from the denominator: $f(x)=\frac{10}{x(\ln(x)- 1)}$. We use the integral test. Let $u = \ln(x)-1$, then $du=\frac{1}{x}dx$. When $x = 2$, $u=\ln(2)-1$ and as $x\rightarrow\infty$, $u\rightarrow\infty$.

Step2: Evaluate the improper - integral

The integral $\int_{2}^{\infty}\frac{10}{x(\ln(x)-1)}dx=10\int_{\ln(2)-1}^{\infty}\frac{1}{u}du$. The antiderivative of $\frac{1}{u}$ is $\ln|u|$. So, $10\int_{\ln(2)-1}^{\infty}\frac{1}{u}du=10\lim_{t\rightarrow\infty}\int_{\ln(2)-1}^{t}\frac{1}{u}du=10\lim_{t\rightarrow\infty}(\ln(t)-\ln(\ln(2)-1))=\infty$.

Answer:

The series diverges.