An alternating series [tex]\sum\limits_n(-1)^na_n[/tex] converges if [tex]|(-1)^na_n|=|a_n|[/tex] is monotonic and [tex]a_n\to0[/tex] as [tex]n\to\infty[/tex]. Here [tex]a_n=\dfrac1{\ln(n+1)}[/tex].
Let [tex]f(x)=\ln(x+1)[/tex]. Then [tex]f'(x)=\dfrac1{x+1}[/tex], which is positive for all [tex]x>-1[/tex], so [tex]\ln(x+1)[/tex] is monotonically increasing for [tex]x>-1[/tex]. This would mean [tex]\dfrac1{\ln(x+1)}[/tex] must be a monotonically decreasing sequence over the same interval, and so must [tex]a_n[/tex].
Because [tex]a_n[/tex] is monotonically increasing, but will still always be positive, it follows that [tex]a_n\to0[/tex] as [tex]n\to\infty[/tex].
So, [tex]\sum\limits_n(-1)^na_n[/tex] converges.