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.

ln(2)/2 + ln(3)/3+ ln(4)/4+ ln(5)/5+ ln(6)/6+ . . . the problem is when using the intergal test its increasing from ln(2)/2 to ln(3)/3 and then decreses from onward....im not sure why its not working?

Answer 1

@wio, this seems outta my league. Can you help this furry fellow out? Fur-rious I get it!

Answer 2

additional info, from 1 to infinity and the formula can also be seen as ln(x+1)/(x+1) i know on the intergal test its positive, continues, but its not decreasing from the first term to the second term...thats the only problem i have with it.... its calc 2 any help :/

Answer 3

The integral test would be showing that \[
\int_0^\infty \frac{\ln(x)}{x}dx
\]converges, no?
Are you having trouble with this integral?

Answer 4

Or rather:
\[\large
\int_2^\infty \frac{\ln(x)}{x}dx
\]Since we start at \(2\)

Answer 5

Hmmm, right it needs to be monotonic....

My suggestion is to just just chop of the first term...
If a series converges, adding a finite term to it doesn't make it stop converging.

Answer 6

\[\large
\sum_{n=2}^\infty \frac{\ln(n)}{n} =\frac{\ln(2)}{2} +\sum_{n=3}^\infty \frac{\ln(n)}{n}
\]Since \(\ln(2)/2\) is finite, then we know either both series converge or both diverge.

Answer 7

yup i got it on my own :) thanks for the answers though, i integrated it to confirm if it diverges which i got \[(\ln (x+1)^2)/2 \] and plugged in infinity and 1 for x and got it went to infinity, therefore \[\sum_{n=1}^{\infty}\]ln(x+1)/(x+1) diverges.