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