# How to prove that $f(x) - f(x-1)$ approaches $\frac{\log_{10}(10)}{\log_{10}(e)}$?

Let $$f(x) = \sum_{n=1}^{10^x}\frac{1}{n}$$

I noticed that as x approaches $\infty$, $f(x) - f(x - 1) \approx 2.3025$. After a bit of experimenting, I found that $2.3025... = \frac{\log_{10}(10)}{\log_{10}(e)}$

How can I prove that as x approaches $\infty$, $f(x) - f(x - 1)$ approaches $\frac{\log_{10}(10)}{\log_{10}(e)}$?

We conclude from here that $$f(x)\approx\ln 10^x + \gamma$$ Subtracting, we get $$f(x)-f(x-1)\approx \ln 10^x - \ln 10^{x-1}=x\ln 10 - (x-1)\ln 10=\ln 10$$ Lastly, we note that $$\frac{\log_{10}10}{\log_{10}e}=\frac{1}{\frac{\ln e}{\ln 10}}=\ln 10$$