Prove that the series is bounded Good afternoon everybody! I have the following problem:
" Prove that the series $s_n=1+\frac{1}{2}+...+\frac{1}{n}-\ln n$ is decreasing and bounded, therefore convergent". 
In order to prove it is decreasing, I must prove that $s_n-s_{n+1} >0$. But $s_n- s_{n+1}=[1+\frac{1}{2}+...+\frac{1}{n}-\ln n]$-$[1+\frac{1}{2}+...+\frac{1}{n}+\frac{1}{n+1}-\ln (n+1)]$= $-\ln n-\frac{1}{n+1}+\ln (n+1)$=$\ln (\frac{n+1}{n})-\frac{1}{n+1}$=$\ln (1+\frac{1}{n})-\frac{1}{n+1}$. If we take the interval $[1,1+\frac{1}{n}]$, the minimum of the function $f(x)=\frac{1}{x}$ is $\frac{n}{n+1}$, hence $\ln(1+\frac{1}{n})=\int_1^{1+\frac{1}{n}} \frac{dx}{x}>\frac{1}{n}\frac{n}{n+1}=\frac{1}{n+1}$. It implies that $s_n-s_{n+1}=\ln(1+\frac{1}{n})-\frac{1}{n+1}>\frac{1}{n+1}-\frac{1}{n+1}=0$.
It shows that the series is monotone. But how to prove the series is also bounded? Thank you!

Note that ${1 \over n+1} \le { 1 \over t } \le {1 \over n}$ for $t \in [n,n+1]$, and $\ln({n+1 \over n}) = \int_n^{n+1} {dt \over t}$, hence ${1 \over n+1} \le \ln({n+1 \over n}) \le {1 \over n}$.
This gives $s_n -s_{n+1} \ge 0$ as you showed above.
Using the other side of the bound and the fact that $\ln$ is non-decreasing we have $s_n = 1+{1 \over 2} +...+{1 \over n} - \ln n \ge \ln({2 \over 1}) + \ln({3 \over 2})+...+\ln({n+1 \over n}) -\ln n = \ln(n+1)-\ln n \ge 0$.
