We will use the alternating series test, which says that for some series #sum(-1)^na_n#, the series converges if #a_n# is decreasing and #lim_(nrarroo)a_n=0#.
We have the series #sum_(n=1)^oo((-1)^(n+1)ln(n+1))/(n+1)#. We see that #(-1)^(n+1)# is the alternating portion so our sequence in question is #a_n=ln(n+1)/(n+1)#.
In order to determine if the series is convergent, we need to determine if the criteria are true.
We can see that #ln(n+1)/(n+1)# is decreasing by noting that #n+1# will increase faster than #ln(n+1)#, so the fraction will get smaller.
You could also graph this, or show that #a_(n-1)/a_n<1#, which means that the current term is greater than the previous term. If we want, we can show that #(ln(n)/(n))/(ln(n+1)/(n+1))<1#.
The second criterion is that #lim_(nrarroo)ln(n+1)/(n+1)=0#. We can do this by again recognizing than #n# grows far faster than #ln(n)#.
We can also use L'Hôpital's rule to find the limit: #lim_(nrarroo)ln(n+1)/(n+1)=lim_(nrarroo)(1/(n+1))/1=0#.
Thus, #sum_(n=1)^oo((-1)^(n+1)ln(n+1))/(n+1)# is a convergent series.
