OMO (2)

Note that $\binom{n+100}{100}^{-1} \; = \; \frac{n! 100!}{(n+100)!} \; = \; \frac{\Gamma(n+1)\Gamma(101)}{\Gamma(n+101)} \; =\; 100 B(n+1,100) \; = \; 100\int_0^1 x^n (1-x)^{99}\,dx$ and so, since $\sum_{n=1}^\infty H_n x^n \; = \; -\frac{\ln(1-x)}{1-x} \hspace{2cm} |x| < 1$ we deduce that $\sum_{n=1}^\infty H_n \binom{n+100}{100}^{-1} \; = \; -100\int_0^1 \ln(1-x) (1-x)^{98}\,dx \; = \; -100\int_0^1 x^{98} \ln x\,dx \; = \; \frac{100}{99^2}$ after integration by parts. This makes the answer $99^2 + 100 = \boxed{9901}$ .