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

\(\frac{\ln2}{2}+\frac{\ln3}{3}+\frac{\ln4}{4}+\frac{\ln5}{5}+\frac{\ln6}{6}+...\)

\(=\sum_{n=2}^\infty\frac{\ln(n)}{n}\)

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