Prove $16\lt\sum_{k=1}^{80}\frac{1}{\sqrt{k}}\lt17$ These are questions I’m stuck on:
(i) Prove for $k\in \mathbb{R}^+:\frac{2}{\sqrt{k+1}+\sqrt{k}}\lt\frac{1}{\sqrt{k}}$
(ii) Prove $16\lt\sum_{k=1}^{80}\frac{1}{\sqrt{k}}\lt17$
I did the first one but I just equated them until I got $\sqrt{k+1}\gt\sqrt{k}$
Which I think I did something wrong because I assumed I was supposed to use induction.
For the second one I am just completely stuck, I tried to convert 16 and 17 into summations with the same limits and got
$\sum_{k=1}^{80}\frac{16k}{3240}=16$
$\sum_{k=1}^{80}\frac{17k}{3240}=17$
But after that I just didn’t know what to do.
 A: As regards (i) it is easier to prove the inequality directly without using induction.
Hint for (ii). Note that
$$\frac{1}{\sqrt{k+1}+\sqrt{k}}=\frac{1}{\sqrt{k+1}+\sqrt{k}}\cdot \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k+1}-\sqrt{k}}=\sqrt{k+1}-\sqrt{k}.$$
Therefore the following sum is telescopic (have you ever encountered such kind of sum?),
$$\sum_{k=1}^{n-1}\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{n}-\sqrt{1}=\sqrt{n}-1.$$
It should be easy now to prove the lower bound.
As regards the upper bound show first that for $k\geq 1$,
$$\frac{1}{\sqrt{k}}<\frac{2}{\sqrt{k-1}+\sqrt{k}}=2(\sqrt{k}-\sqrt{k-1}).$$
Can you take it from here?
A: We have
$$\int_1^{81}\frac{1}{\sqrt{x}}\:dx=\left[  2\sqrt{x}\right]_1^{81}=18-2=16$$
Using a Riemann sum with circumscribed rectangles will  overestimate  the integral. We use rectangles of width $1$. Since the function is decreasing, the heights of the rectangles will be based on function values at left endpoints.
This gives $$16 =\int_1^{81}\frac{1}{\sqrt{x}}\:dx < \sum_{k=1}^{80} \frac{1}{\sqrt{k}}$$
Next we underestimate the integral using a Riemann sum with inscribed rectangles. We get
$$\sum_{k=2}^{81} \frac{1}{\sqrt{k}}< \int_1^{81}\frac{1}{\sqrt{x}}\:dx=16$$
Adding $1$ to both sides of this inequality gives
$\sum_{k=1}^{81} \frac{1}{\sqrt{k}}< 17$
But then $$\sum_{k=1}^{80} \frac{1}{\sqrt{k}}< \sum_{k=1}^{81} \frac{1}{\sqrt{k}}<17$$
Putting it all together gives the desired inequality.
