It will diverge. Consider the Nth partial sum,
[tex]\displaystyle\sum_{n=1}^N\ln\frac n{n+5}=\sum_{n=1}^N(\ln n-\ln(n+5))[/tex]
= (ln1 - ln6) + (ln2 - ln7) + (ln3 - ln8) + (ln4 - ln9) + (ln5 - ln10)
+ (ln6 - ln11) + (ln7 - ln12) + ...
+ (ln(N-5) - lnN) + (ln(N-4) - ln(N+1)) + (ln(N-3) - ln(N+2)) + (ln(N-2) - ln(N+3)) + (ln(N-1) - ln(N+4))
+ (lnN - ln(N+5))
Cancellation among terms leaves us with
= ln1 + ln2 + ln3 + ln4 + ln5 - ln(N+1) - ln(N+2) - ln(N+3) - ln(N+4) - ln(N+5)
We can write this as
[tex]\ln(1\cdot2\cdot3\cdot4\cdot5)-\ln((N+1)\cdot(N+2)\cdot(N+3)\cdot(N+4)\cdot(N+5))[/tex]
[tex]\implies\displaystyle\sum_{n=1}^N\ln\frac n{n+5}=\ln\frac{120}{(N+5)(N+4)(N+3)(N+2)(N+1)}[/tex]
Then as [tex]N\to\infty[/tex], the argument of the logarithm is 0, so the logarithm itself approaches -infinity. Then the limit of the series is non-zero, so it must diverge.