Show that $1 + 1/2 + 1/3 + .... + 1/n - \ln (n)$ converges.

Question

Show that $1 + 1/2 + 1/3 + .... + 1/n - \ln (n)$ converges. I tried to prove it using the comparison test.

First I made it into a sum of 1 + $\sum_{2}^{n} \frac{1}{m} -\ln(\frac{m}{m-1})$ and tried to do comparison with $\frac{1}{m^2}$. From $|\frac{1}{m} - \ln(\frac{m}{m-1})| < \frac{1}{m^2}$, I managed to prove $\frac{1}{m} - \ln(\frac{m}{m-1}) < \frac{1}{m^2}$ but I am unable to show that $-\frac{1}{m} + \ln(\frac{m}{m-1}) < \frac{1}{m^2}$. I am wondering how to prove the last inequality if its possible.

Answer 1

If you don't accuracy, just note that $\log n = \int_{1}^{n} \frac{dt}{t}$ and $\sum_{k=1}^{n}\frac{1}{k} < \int_{1}^{n+1}\frac{dt}{t}$. Once you subtract, you get $\int_{n}^{n+1}\frac{dt}{t} = \log (1+ \frac{1}{n})$ which converges to $0$.