Greatest integer that is less than or equal to $\sum_{n=1}^{100}\frac{1}{\sqrt{n}}$

Question

Basically I want to find the GIF of $\sum_{n=1}^{100}\frac{1}{\sqrt{n}}$

I have no idea how to do it. Can anyone give me a hint for solving this. thanks!

Answer 1

A bit late this answer but i thought it could be worth mentioning it.

You can also solve your problem without integrals in an elementary way using

$$\sqrt{n+1}-\sqrt n = \frac 1{\sqrt{n+1}+\sqrt n}$$

Hence, you have the estimations

$$\frac 1{2\sqrt{n+1}}<\sqrt{n+1}-\sqrt n< \frac 1{2\sqrt{n}}$$

Now, you apply this to your sum:

\begin{eqnarray*}2\sum_{n=1}^{100}(\sqrt{n+1}-\sqrt n) & < & \sum_{n=1}^{100}\frac 1{\sqrt n} = 1+ \sum_{n=\color{blue}{1}}^{\color{blue}{99}}\frac 1{\sqrt{\color{blue}{n+1}}} \\ & < & 1+2\sum_{n={1}}^{{99}}(\sqrt{n+1}-\sqrt n) \end{eqnarray*}

Now, only two telescoping sums are to be evaluated:

$$18 < 2(\sqrt{101}-1) <\sum_{n=1}^{100}\frac 1{\sqrt n}< 1+2(10-1) = 19$$

Answer 2

We can do Integral comparison test:

$\displaystyle \tag*{} \begin{align} \int \limits _{1}^{101} \dfrac{1}{\sqrt{n}}\ \mathrm dn <\sum \limits _{n=1}^{100} \dfrac{1}{\sqrt{n}} &< 1+\int \limits_{1}^{100} \dfrac{1}{\sqrt{n}} \ \mathrm dn \\\\ 18.1 <\sum \limits _{n=1}^{100} \dfrac{1}{\sqrt{n}} &< 19 \end{align}$

So, we have:

$\displaystyle \tag*{} \left \lfloor \sum \limits _{n=1}^{100} \dfrac{1}{\sqrt{n}} \right \rfloor = 18$