If f(n)= 1 + 2 + 3 +∙∙∙+(n+1) and $$g(n)$$ = $$\sum_{k=1}^{k=n}\frac{1}{f(k)}$$ then the least value of $$n$$ for which $$g(n)$$ exceeds the value $$\frac{99}{100}$$ is ____________.

$$f(n)=1+2+3+.....+(n+1)$$

So, $$f(n)=\dfrac{\left(n+1\right)\left(n+2\right)}{2}$$

or, $$\dfrac{1}{f(n)}=\dfrac{2}{\left(n+1\right)\left(n+2\right)}$$

or, $$\dfrac{1}{f(n)}=2\left[\dfrac{1}{n+1}-\dfrac{1}{n+2}\right]$$

So, $$\dfrac{1}{f(1)}=2\left[\dfrac{1}{2}-\dfrac{1}{3}\right]$$

$$\dfrac{1}{f\left(2\right)}=2\left[\dfrac{1}{3}-\dfrac{1}{4}\right]$$

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$$\dfrac{1}{f\left(n\right)}=2\left[\dfrac{1}{n+1}-\dfrac{1}{n+2}\right]$$

Adding all the terms,

$$g(n)$$ =  $$\sum_{k=1}^{k=n}\dfrac{1}{f(k)}$$ = $$2\left[\dfrac{1}{2}-\dfrac{1}{n+2}\right]$$

or, $$g\left(n\right)=\dfrac{2\left(n+2-2\right)}{\left(n+2\right)2}$$

or, $$g\left(n\right)=\dfrac{n}{n+2}$$

Now, $$g\left(n\right)>\ \dfrac{99}{100}$$

or, $$\dfrac{n}{n+2}>\ \dfrac{99}{100}$$

or, $$100n>\ 99n+198$$

or, $$n>198$$

So least value of $$n$$ is 199