Answer: Let the total area of the unit-width rectangles be
\(\displaystyle \frac{1}{\sqrt{8}}+\frac{1}{\sqrt{15}}+\cdots+\frac{1}{\sqrt{n^2+2n}}=\sum_{r=2}^{n}\frac{1}{\sqrt{r^2+2r}}.\)
Since the curve \(y=\dfrac{1}{\sqrt{x^2+2x}}\) is decreasing for \(x\gt 0\), these rectangles lie below the curve between \(x=1\) and \(x=n\). Hence
\(\displaystyle \sum_{r=2}^{n}\frac{1}{\sqrt{r^2+2r}}\lt \int_1^n \frac{1}{\sqrt{x^2+2x}}\,dx.\)
Now
\(x^2+2x=(x+1)^2-1,\)
so
\(\displaystyle \int_1^n \frac{1}{\sqrt{x^2+2x}}\,dx=\int_1^n \frac{1}{\sqrt{(x+1)^2-1}}\,dx\)
\(\displaystyle =\left[\cosh^{-1}(x+1)\right]_1^n\)
\(\displaystyle =\left[\ln\bigl(x+1+\sqrt{x^2+2x}\bigr)\right]_1^n\)
\(\displaystyle =\ln\left(n+1+\sqrt{n^2+2n}\right)-\ln(2+\sqrt3).\)
Also, the missing first term is
\(\displaystyle \frac{1}{\sqrt{1^2+2\cdot1}}=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}.\)
Therefore
\(\displaystyle \sum_{r=1}^{n} \frac{1}{\sqrt{r^2+2r}}\lt \ln\left(n+1+\sqrt{n^2+2n}\right)+\frac{\sqrt3}{3}-\ln(2+\sqrt3).\)
The curve is \(y=\dfrac{1}{\sqrt{x^2+2x}}\). For \(x\gt 0\), this function is decreasing, so the rectangles shown lie below the curve.
Each rectangle has width \(1\), and their heights are
\(\displaystyle \frac{1}{\sqrt{8}},\ \frac{1}{\sqrt{15}},\ \ldots,\ \frac{1}{\sqrt{n^2+2n}},\)
so the total area of the \((n-1)\) rectangles is
\(\displaystyle \frac{1}{\sqrt{8}}+\frac{1}{\sqrt{15}}+\cdots+\frac{1}{\sqrt{n^2+2n}}=\sum_{r=2}^{n}\frac{1}{\sqrt{r^2+2r}}.\)
Because these rectangles are beneath the curve from \(x=1\) to \(x=n\),
\(\displaystyle \sum_{r=2}^{n}\frac{1}{\sqrt{r^2+2r}}\lt \int_1^n \frac{1}{\sqrt{x^2+2x}}\,dx.\)
To evaluate the integral, complete the square:
\(x^2+2x=(x+1)^2-1.\)
Hence
\(\displaystyle \int_1^n \frac{1}{\sqrt{x^2+2x}}\,dx=\int_1^n \frac{1}{\sqrt{(x+1)^2-1}}\,dx.\)
Using the standard result
\(\displaystyle \int \frac{1}{\sqrt{u^2-1}}\,du=\cosh^{-1}(u)=\ln\bigl(u+\sqrt{u^2-1}\bigr),\)
with \(u=x+1\), we get
\(\displaystyle \int_1^n \frac{1}{\sqrt{(x+1)^2-1}}\,dx=\left[\cosh^{-1}(x+1)\right]_1^n\)
\(\displaystyle =\left[\ln\bigl(x+1+\sqrt{x^2+2x}\bigr)\right]_1^n.\)
Substituting the limits gives
\(\displaystyle \int_1^n \frac{1}{\sqrt{x^2+2x}}\,dx=\ln\left(n+1+\sqrt{n^2+2n}\right)-\ln(2+\sqrt3).\)
Therefore
\(\displaystyle \sum_{r=2}^{n}\frac{1}{\sqrt{r^2+2r}}\lt \ln\left(n+1+\sqrt{n^2+2n}\right)-\ln(2+\sqrt3).\)
Now add the omitted term \(r=1\):
\(\displaystyle \frac{1}{\sqrt{1^2+2\cdot1}}=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}.\)
So
\(\displaystyle \sum_{r=1}^{n}\frac{1}{\sqrt{r^2+2r}}=\frac{\sqrt3}{3}+\sum_{r=2}^{n}\frac{1}{\sqrt{r^2+2r}},\)
and hence
\(\displaystyle \sum_{r=1}^{n}\frac{1}{\sqrt{r^2+2r}}\lt \ln\left(n+1+\sqrt{n^2+2n}\right)+\frac{\sqrt3}{3}-\ln(2+\sqrt3).\)
This is the required result.
