Show that $\displaystyle 16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17$.

Show that $\displaystyle 16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17$.

On my previous post (Show that $\displaystyle 16<\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{80}}<17$), we used the trapezoid rule to prove the upper bound for the given inequality, i.e. $\displaystyle 16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}$. We're then not supposed to use the same method to prove the lower bound simply because the function $y=\dfrac{1}{\sqrt{x}}$ is concave up. No matter how we manipulated that concept, we will only end up with proving the target sum $\displaystyle \sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}$ will be greater than some quantity, not less than. This works against to what we are looking to prove, that is, $\displaystyle \sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17$.

We have to come up with another idea. More often than not, questions ask to estimate the sum are questions of telescoping series.

But, the thing is, how exactly

$\displaystyle \sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\cdots +\dfrac{1}{\sqrt{80}}$

can be a telescoping series in disguise?

Okay, let's start from the very obvious, then work things out in details:

It is very clear that:

$k>k-1$

Taking square root of both sides yields:

$\sqrt{k}>\sqrt{k-1}$

Adding the quantity $\sqrt{k}$ to both sides gives:

$\sqrt{k}+\sqrt{k}>\sqrt{k}+\sqrt{k-1}$

Don't worry, albeit we got a summation here, but, if we turn both sides into fractions, then, when we rationalize the denominator, we would get a difference instead.

$2\sqrt{k}>\sqrt{k}+\sqrt{k-1}$

$\dfrac{2}{\sqrt{k}+\sqrt{k-1}}>\dfrac{1}{\sqrt{k}}$ (Note that $\displaystyle \color{yellow}\bbox[5px,purple]{k\ne 1}$ or the fraction $\dfrac{1}{0}$ would be undefined and then the whole sum will follow suit, becomes undefined as well.

$\dfrac{2}{\sqrt{k}+\sqrt{k-1}}\dfrac{\cdot \sqrt{k}-\sqrt{k-1}}{\cdot \sqrt{k}-\sqrt{k-1}}>\dfrac{1}{\sqrt{k}}$

$\dfrac{2(\sqrt{k}-\sqrt{k-1})}{1}>\dfrac{1}{\sqrt{k}}$

$2(\sqrt{k}-\sqrt{k-1})>\dfrac{1}{\sqrt{k}}$

Remember that we're asked to find $\displaystyle \sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17$

One new condition that we just set, we must satisfy that, that means we have to algebraically manipulate the relation of inequality $\dfrac{2(\sqrt{k}-\sqrt{k-1})}{1}>\dfrac{1}{\sqrt{k}}$ so we could estimate the sum of $\displaystyle \sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}$.

$\displaystyle \sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}=\dfrac{1}{\sqrt{1}}+\sum_{k=2}^{80}\dfrac{1}{\sqrt{k}}$

For $n=2$:  $2(\bcancel{\sqrt{2}}-\sqrt{1})>\dfrac{1}{\sqrt{2}}$

For $n=3$:  $2(\bcancel{\sqrt{3}}-\bcancel{\sqrt{2}})>\dfrac{1}{\sqrt{3}}$

For $n=4$:  $2(\bcancel{\sqrt{4}}-\bcancel{\sqrt{3}})>\dfrac{1}{\sqrt{4}}$

       $\vdots$           $\vdots$                    $\vdots$

For $n=79$:  $2(\bcancel{\sqrt{79}}-\bcancel{\sqrt{78}})>\dfrac{1}{\sqrt{79}}$

For $n=80$:  $2(\sqrt{80}-\bcancel{\sqrt{79}})>\dfrac{1}{\sqrt{80}}$

Therefore

$\displaystyle \begin{align*}\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+\cdots+\dfrac{1}{\sqrt{80}}&=\dfrac{1}{\sqrt{1}}+\sum_{k=2}^{80}\dfrac{1}{\sqrt{k}}\\& < 1+(-2\sqrt{1})+(2\sqrt{80})\\& < 2\sqrt{80}-1 \end{align*}$

Since

$81>80$

$9> \sqrt{80}$

$18> 2\sqrt{80}$

$17> 2\sqrt{80}-1$

We can put together what we just found to see that:

$\displaystyle \sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}=\dfrac{1}{\sqrt{1}}+\sum_{k=2}^{80}\dfrac{1}{\sqrt{k}}< 2\sqrt{80}-1< 17$

In other words, we have proved that

$\displaystyle \sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}< 17$.

And we're done.