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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!

Gary
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longnh
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    Keep the first few terms as they are and estimate the tail via $ 2(\sqrt {2018} - \sqrt m ) < \sum\limits_{k = m}^{2017} {\frac{1}{{\sqrt k }}} < 2(\sqrt {2017} - \sqrt {m - 1} ) $ with a suitable $m$. – Gary Feb 05 '22 at 06:48
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    Similar questions https://math.stackexchange.com/q/540420/1012971 and https://math.stackexchange.com/questions/1826197/floor-sum-of-reciprocal-of-square-root-of-first-50-numbers?noredirect=1 and https://artofproblemsolving.com/community/c4h1892199p12914972 –  Feb 05 '22 at 07:45
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    Thank you @Gary, I've solved this with your help! – longnh Feb 05 '22 at 07:55
  • Thank you @RamanujanXV, I did a search, but they all used integral which is not allowed in solution. – longnh Feb 05 '22 at 07:56
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    @longnh So make an answer based on Gray's help in below. :) – Mikasa Feb 05 '22 at 07:57

1 Answers1

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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 $$\lfloor S\rfloor = 88.$$
Mikasa
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longnh
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  • Same approach as here https://math.stackexchange.com/a/1826218/42969, and that does not use integrals. – Martin R Feb 05 '22 at 08:18