Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series. frac{ln2}{2}+frac{ln3}{3}+frac{ln4}{4}+frac{ln5}{5}+frac{ln6}{6}+...

Series
Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.
$$\frac{\ln2}{2}+\frac{\ln3}{3}+\frac{\ln4}{4}+\frac{\ln5}{5}+\frac{\ln6}{6}+...$$

2021-02-12
We have the given series is,
$$\frac{\ln2}{2}+\frac{\ln3}{3}+\frac{\ln4}{4}+\frac{\ln5}{5}+\frac{\ln6}{6}+...$$
$$=\sum_{n=2}^\infty\frac{\ln(n)}{n}$$
Before getting into the integral test, we must assure two things first : for the integral step to apply to
$$\sum_{n=N}^\infty f(n)$$ we must have $$a_n>0$$ for the given interval and f(n) must be decresing on the same interval.
Both of these are true since
$$\ln(n)>0$$ and $$n>0$$ on $$n\in[2,\infty)$$ so $$\frac{\ln(n)}{n}>0$$ on the same interval.
Furthermore, $$\ln(n)$$ grows slower than the n so we see that $$\frac{\ln(n)}{n}$$ is decresing because n overpowers $$\ln(n)$$ in the numerator.
We can also sjow this by taking derivative of $$\frac{\ln(n)}{n}$$ and showing it's away negative on $$n\in[2,\infty)$$.
So wee see the integral test applies.
The integral test states that if the two if two condition are met,then for $$\sum_{n=N}^\infty f(n)$$, evaluate the improper integral $$\int_N^\infty f(x)dx$$
If the integral converges to a real,finite value,then the series converges.If the integral diverges, then the series does too.
So, we take the integral $$\int_2^\infty\frac{\ln(x)}{x}dx$$
Now by taking the limit as it goes to infinity
$$\int_2^\infty\frac{\ln(x)}{x}dx=\lim_{b\to\infty}\int_2^\infty\frac{\ln(x)}{x}dx$$
letting $$u=\ln(x)$$
so $$du=\frac{1}{x}dx$$
$$=\lim_{b\to\infty}\int_{\ln(2)}^{\ln(b)}udu$$
$$=\lim_{b\to\infty}\left[\frac{1}{2}u^2\right]_{\ln(2)}^{\ln(b)}$$
$$=\lim_{b\to\infty}\frac{1}{2}\ln^2(b)-\frac{1}{2}\ln^2(2)$$
as $$b\to\infty$$, we see that $$\ln(b)\to\infty$$, so,
$$=\infty$$
hence, the integral diverges.
Thus, we see that
$$\sum_{n=2}^\infty\frac{\ln(n)}{n}$$ diverges as well