How to find floor of $ 1+\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ...+\frac{1}{\sqrt{2017}}$?

Question

How to find floor of $S=1+\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ...+\frac{1}{\sqrt{2017}}$?

With $$\frac{1}{\sqrt{i}} > 2(\sqrt{i+1} -\sqrt{i})$$ and $$\frac{1}{\sqrt{i}} < 2(\sqrt{i} -\sqrt{i-1}),$$ I'm able to get to $$2 \sqrt{n+1}-2 < \sum_{i=1}^n \frac{1}{\sqrt{i}} < 2 \sqrt{n}-1.$$

Given $n=2017$, the result is $88.37282...$ (using Wolfram), so the floor should be $88$.
But I just get to $87.844<S<88.822$, and therefore I don't get the exact floor.
With ALGEBRA ONLY (no integral, limit, Harmonic Series), could you please help me to solve this problem?
Thank you!

Answer 1

Thanks to @Gary's help, I'm able to solve this problem. Thank you!
With $k≥1$, we have:
$$ \frac{2}{\sqrt{k} + \sqrt{k+1}} < \frac{2}{\sqrt{k} + \sqrt{k}} < \frac{2}{\sqrt{k} + \sqrt{k-1}} $$ So $$ \frac{2}{\sqrt{k} + \sqrt{k+1}} < \frac{1}{\sqrt{k}} < \frac{2}{\sqrt{k} + \sqrt{k-1}} $$ and so $$ 2(\sqrt{k+1} - \sqrt{k}) < \frac{1}{\sqrt{k}} < 2(\sqrt{k} - \sqrt{k-1})$$

• With $k=2$: $ 2(\sqrt{3} - \sqrt{2}) < \frac{1}{\sqrt{2}} < 2(\sqrt{2} - \sqrt{1})$
• With $k=3$: $ 2(\sqrt{4} - \sqrt{3}) < \frac{1}{\sqrt{3}} < 2(\sqrt{3} - \sqrt{2})$
  ...
• With $k=2017$: $ 2(\sqrt{2018} - \sqrt{2017}) < \frac{1}{\sqrt{2017}} < 2(\sqrt{2017} - \sqrt{2016})$
  So we have: $1 + 2(\sqrt{2018} - \sqrt{2}) < S < 1 + 2(\sqrt{2017} - 1)$
  And therefore $88.016<S<88.822$ which implies that $$\lfloor S\rfloor = 88.$$