Is it a Board Level Problem?!

We have $\sum_{r=0}^n (-1)^r \binom{n}{r}\frac{1}{(1 + n\ln10)^r} \; =\; \left(1 - \frac{1}{1 + n\ln10}\right)^n \; = \; \left(\frac{n\ln10}{1 + n\ln10}\right)^n$ while $\begin{aligned}\sum_{r=0}^n (-1)^r\binom{n}{r} \frac{r \ln10}{(1 + n\ln10)^r} & = \; \sum_{r=1}^n (-1)^r n \binom{n-1}{r-1} \frac{\ln10}{(1 + n\ln10)^r} \; = \; n\ln10\sum_{r=0}^{n-1}(-1)^{r-1}\binom{n-1}{r} \frac{1}{(1 + n\ln10)^{r+1}} \\ & = \; -\frac{n \ln10}{1 + n\ln10} \left(1 - \frac{1}{1 + n\ln10}\right)^{n-1} \; = \; -\left(\frac{n\ln10}{1 + n\ln10}\right)^n \end{aligned}$ Thus $S = 0$ .