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.

 A: To show that $D_n$ is bounded above observe that
$$
D_n \leq 1 + \int_1^n \frac{1}{x}\,dx - \int_1^{n+1} \frac{1}{x}\,dx = 1 + \ln\left(\frac{n}{n+1}\right) \leq 1 + \ln(1) = 1
$$
You may obtain this inequality as follows. Observe that $f(x) = 1/x$ is monotonically decreasing on $[1,\infty)$. Thus, $f(k) \leq f(x) \leq f(k-1)$ on the interval $[k-1,k]$ for all integers $k \geq 2$. It follows that
$$
f(k) = \int_{k-1}^k f(k)\,dx \leq \int_{k-1}^k f(x)\,dx \implies \sum_{k=2}^n \frac{1}{k} \leq \int_1^n \frac{1}{x}\,dx
$$
Adding $1$ to both sides of the inequality gives
$$
\sum_{k=1}^n \frac{1}{k} \leq 1 + \int_1^n \frac{1}{x}\,dx
$$
This is essentially just the proof of the integral test for convergence of infinte series.
A: 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
$$D_n \leq \left|\sum_{k=2}^{n+1} \frac{1}{k} - \sum_{k=1}^{n}\frac{1}{k}\right|=\left|\frac{1}{n+1}-1\right|\leq \frac{1}{n+1}+1 \leq 2.$$
A: $D_{n+1}-D_n=\{\sum_{n=1}^\infty \frac{1}{n+1}-\int_1^{n+2}\frac{1}{x}\}-\{\sum_{n=1}^\infty \frac{1}{n+2}-\int_1^{n+1}\frac{1}{x}\}$
$=\frac{1}{2}-\{\int _{n+1}^{n+2}\frac{1}{x} \}=\frac{1}{2}-\ln{\frac{n+2}{n+1}}$
$=\frac{1}{2}+\ln{\frac{n+1}{n+2}}>0$
A: Why it is increasing:

Why it is bounded: (just note the Red is bounded)

(Dissatisfying that to show the Red is bounded still needs a little calculus. I'll be glad to see any idea that can de-calculise the proof of boundedness.)
