An Old Problem Came To Mind – Mattia Giuri's bizarre blog

Two years ago I participated in an online math competition and the problem which was considered as the most difficult was this one:

$x = \sum_{n = 1}^{10000} \frac{1}{\sqrt{n}}$

Find

$\lfloor x \rfloor$

Now the trick is finding upper and lower bounds for such quantity. Notice:

$\frac{1}{2\sqrt{n}} = \frac{1}{\sqrt{n}+\sqrt{n}} > \frac{1}{\sqrt{n+1}+\sqrt{n}} = \sqrt{n+1} - \sqrt{n}$

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

Similarly

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

$\Rightarrow \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})$

If we sum up this relationship for all values we’ll found out it’s telescoping, so we’re left with:

$2\sqrt{n+1} - 2\sqrt{2} + 1 < \sum_{k = 1}^{n} \frac{1}{\sqrt{k}} < 2\sqrt{n} - 2 + 1$

Hence for $n=10000$ we get

$198 + \epsilon < x < 199$

Hence, we get that the value we were looking for is

$198$