Answer: (a) \(\displaystyle \sum_{r=1}^{n} \frac{1}{\sqrt{r}}\,\mathrm e^{\sqrt r} \lt \left(2+\frac{1}{\sqrt n}\right)\mathrm e^{\sqrt n}-2\mathrm e\).

(b) A lower bound is \(\displaystyle \sum_{r=1}^{n} \frac{1}{\sqrt{r}}\,\mathrm e^{\sqrt r} \gt 2\mathrm e^{\sqrt n}-\mathrm e\).

Let \(f(x)=\dfrac{1}{\sqrt x}\,\mathrm e^{\sqrt x}\) for \(x\ge 1\).

Since \(f(x)\) is increasing for \(x\ge 1\), left-endpoint rectangles of width 1 lie below the curve, while right-endpoint rectangles lie above it.

(a) Using the \(n-1\) rectangles with heights \(f(1),f(2),\dots,f(n-1)\), the total area is

\(\displaystyle \sum_{r=1}^{n-1} \frac{1}{\sqrt r}\,\mathrm e^{\sqrt r}.\)

Hence

\(\displaystyle \sum_{r=1}^{n-1} \frac{1}{\sqrt r}\,\mathrm e^{\sqrt r} \lt \int_1^n \frac{1}{\sqrt x}\,\mathrm e^{\sqrt x}\,dx.\)

Now

\(\displaystyle \int \frac{1}{\sqrt x}\,\mathrm e^{\sqrt x}\,dx = 2\mathrm e^{\sqrt x},\)

so

\(\displaystyle \int_1^n \frac{1}{\sqrt x}\,\mathrm e^{\sqrt x}\,dx = \left[2\mathrm e^{\sqrt x}\right]_1^n = 2\mathrm e^{\sqrt n}-2\mathrm e.\)

Therefore

\(\displaystyle \sum_{r=1}^{n-1} \frac{1}{\sqrt r}\,\mathrm e^{\sqrt r} \lt 2\mathrm e^{\sqrt n}-2\mathrm e.\)

Adding the \(r=n\) term gives

\(\displaystyle \sum_{r=1}^{n} \frac{1}{\sqrt r}\,\mathrm e^{\sqrt r} \lt 2\mathrm e^{\sqrt n}-2\mathrm e+\frac{1}{\sqrt n}\,\mathrm e^{\sqrt n},\)

so

\(\displaystyle \sum_{r=1}^{n} \frac{1}{\sqrt r}\,\mathrm e^{\sqrt r} \lt \left(2+\frac{1}{\sqrt n}\right)\mathrm e^{\sqrt n}-2\mathrm e.\)

(b) Using the rectangles with heights \(f(2),f(3),\dots,f(n)\), the total area is

\(\displaystyle \sum_{r=2}^{n} \frac{1}{\sqrt r}\,\mathrm e^{\sqrt r}.\)

These rectangles lie above the curve, so

\(\displaystyle \sum_{r=2}^{n} \frac{1}{\sqrt r}\,\mathrm e^{\sqrt r} \gt \int_1^n \frac{1}{\sqrt x}\,\mathrm e^{\sqrt x}\,dx = 2\mathrm e^{\sqrt n}-2\mathrm e.\)

Now add the missing \(r=1\) term, namely \(\mathrm e\):

\(\displaystyle \sum_{r=1}^{n} \frac{1}{\sqrt r}\,\mathrm e^{\sqrt r} \gt 2\mathrm e^{\sqrt n}-2\mathrm e+\mathrm e = 2\mathrm e^{\sqrt n}-\mathrm e.\)