<s>[INST] Show a sequence is increasing 
How can you show that :
$$D_n=\sum_{k=1}^{n } \frac{1}{k}-\int_{1}^{n+1} \frac{1}{x} \ dx $$ is increasing and bounded (and hence convergent). I'm having trouble.

 [/INST] Assuming you meant
$$D_n=\sum_{k=1}^n \frac{1}{k}-\int_1^{n+1}\frac{1}{x}\ dx$$
we have
\begin{align*}
D_{n+1}-D_n &= \left(\sum_{k=1}^{n+1} \frac{1}{k}-\int_1^{n+2}\frac{1}{x}\ dx\right)-\left(\sum_{k=1}^n \frac{1}{k}-\int_1^{n+1}\frac{1}{x}\ dx\right)\\
&= \frac{1}{n+1}-\int_1^{n+2}\frac{1}{x} \ dx+\int_1^{n+1}\frac{1}{x}\ dx\\
&=\frac{1}{n+1}-\int_{n+1}^{n+2} \frac{1}{x} \ dx.
\end{align*}
Now on $[n+1,n+2]$ we have $x \geq n+1$ so $\frac{1}{x} \leq \frac{1}{n+1}$. Hence $\int_{n+1}^{n+2} \frac{1}{x} \ dx \leq \int_{n+1}^{n+2} \frac{1}{n+1} \ dx = \frac{1}{n+1}$. In fact the inequality is strict because we only have $\frac{1}{x}=\frac{1}{n+1}$ when $x=n+1$. This shows $D_{n+1}-D_n>0$ so $D_n$ is increasing.
To see that $D_n$ is bounded, note that $\sum_{k=1}^n \frac{1}{k}$ is a left Riemann sum for $\int_1^{n+1} \frac{1}{x} \ dx$. Hence $D_n$ is the difference between an integral and left Riemann sum, so $D_n$ is bounded by $|R_n-L_n|$ where $R_n$ is a right Riemann sum and $L_n$ is a left Riemann sum for $\int_1^{n+1} \frac{1}{x} \ dx$. One particular choice of left and right Riemann sums gives
$$D_n \leq \left|\sum_{k=2}^{n+1} \frac{1}{k} - \sum_{k=1}^{n}\frac{1}{k}\right|=\left|\frac{1}{n+1}-1\right|\leq \frac{1}{n+1}+1 \leq 2.$$
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