For each positive integer n, let
{y_n}= 1 + 1/2 +...+1/n- the integral of (1/x) from 1 to n.
Prove that the sequence {y_n} converges.
Proof of monotone decreasing: $\displaystyle y_{n-1}-y_n=\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}+\int_1^n\frac{1}{x}\,dx-\int_1^{n-1}\frac{1}{x}\,dx$ $\displaystyle =\int_{n-1}^n\frac{1}{x}\,dx-\frac{1}{n}\geq\int_{n-1}^n\frac{1}{n}\,dx-\frac{1}{n}=0$
To prove that $\displaystyle \{y_n\}$ is bounded below by $\displaystyle 0$, if $\displaystyle P$ is the partition $\displaystyle \{x_1=1,x_2=2,...,x_n=n\}$, and $\displaystyle f(x)=\frac{1}{x}$, the upper sum $\displaystyle U(P,f)$ of the integral of $\displaystyle f(x)$ is $\displaystyle \sum_{k=1}^{n-1}\frac{1}{x_k}(x_{k+1}-x_k)=\sum_{k=1}^{n-1}\frac{1}{k}$. Since $\displaystyle \sum_{k=1}^n\frac{1}{k}>U(P,f)>\int_1^n\frac{1}{x} \,dx$, it follows that $\displaystyle y_n>0$ for all $\displaystyle n$.
Since $\displaystyle \{y_n\}$ is a monotonically decreasing sequence bounded below, it converges.
This is actually a famous limit and converges to the Euler-Mascheroni Constant, $\displaystyle \gamma$.
$\displaystyle \lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k}-\ln(n)\right)=\gamma\approx0.577$
You're welcome.