Question

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}+...

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ANSWERED
asked 2021-02-11
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.
\(\frac{\ln2}{2}+\frac{\ln3}{3}+\frac{\ln4}{4}+\frac{\ln5}{5}+\frac{\ln6}{6}+...\)

Expert Answers (1)

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
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