If \(f(n)= 1 + 2 + 3 +\cdots+(n+1) \) and \(g(n)= \sum_{k=1}^{k=n} \dfrac{1}{f(k)}\), then the least value of \(n\) for which \(g(n)\) exceeds the value \(\dfrac{99}{100}\) is:

The least value of $n$ is 199. Since $f(k)=\tfrac{(k+1)(k+2)}2$, its reciprocal is $\tfrac2{k+1}-\tfrac2{k+2}$, and the sum telescopes to $1-\tfrac2{n+2}$. Exceeding $\tfrac{99}{100}$ requires $n+2>200$, hence $n>198$. At 198 the sum merely equals the threshold.