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I need help proving that the following statement is false:

$$ 399 < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{40000}} < 400. $$

I tried to bound the summation between two integrals

$$ \text{1)} \int_1^{40001} \frac{2}{\sqrt{x}} \qquad \text{and} \qquad \text{2)} \int_0^{40000} \frac{2}{\sqrt{x}} $$

and have convinced myself that the statement is true, but this is not the correct answer as per my textbook.

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  • $\begingroup$ the sum is very close to $ 2 \sqrt{40000} + \zeta \left( \frac{1}{2} \right) $ where $ \zeta \left( \frac{1}{2} \right) \approx -1.4603545 $ $\endgroup$
    – Will Jagy
    Aug 6, 2021 at 4:33
  • $\begingroup$ Is there something wrong with what I did? $\endgroup$
    – Ilovemath
    Aug 6, 2021 at 5:54
  • $\begingroup$ @WillJagy. Tahing into account our last exchange, this is $H_n^{\left(\frac{1}{2}\right)}$ and its expansion is $2 \sqrt{n}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2\sqrt n}+\cdots$ $\endgroup$ Aug 6, 2021 at 5:57
  • $\begingroup$ It might be worth visualizing the functions that you are using to bound the sum, that might give you some ideas on how to further bound it. Also, have you tried writing a simple code to sum all that? If you do sou you will "unconvince" yourself. Just because the bound that you got allow the values of the sum to be in that interval it doesn't mean they will be. Unless of course the bound (interval) you got is contained (or equal) to that interval given by the inequalities of the statement. $\endgroup$
    – jDAQ
    Aug 6, 2021 at 7:06
  • $\begingroup$ @ClaudeLeibovici thanks for your replies; I will try to work up how Robjohn got to his simple statement. $\endgroup$
    – Will Jagy
    Aug 6, 2021 at 17:50

2 Answers 2

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Remarks: Actually my solution is not nice. A very nice solution is given by @Jack D'Aurizio in Bounding sum of reciprocals of the square roots of the first N positive integers

We have \begin{align*} S &= \sum_{k=1}^{40000} \frac{2}{\sqrt{k} + \sqrt{k + 1}} + \sum_{k=1}^{40000} \left(\frac{1}{\sqrt{k}} - \frac{2}{\sqrt{k} + \sqrt{k + 1}}\right)\\ &= \sum_{k=1}^{40000}2(\sqrt{k + 1} - \sqrt{k}) + \sum_{k=1}^{40000} \frac{1}{\sqrt{k}\, (\sqrt{k + 1} + \sqrt{k})^2}\\ &= 2(\sqrt{40001} - \sqrt{1}) + \sum_{k=1}^{40000} \frac{1}{\sqrt{k}\, (\sqrt{k + 1} + \sqrt{k})^2}\\ &\le 2(\sqrt{40001} - \sqrt{1}) + \sum_{k=1}^{40000} \frac{1}{\sqrt{k}\, (\sqrt{k} + \sqrt{k})^2}\\ &= 2(\sqrt{40001} - \sqrt{1}) + \frac{1}{4}\sum_{k=1}^{40000} \frac{1}{k^{3/2}}\\ &= 2(\sqrt{40001} - \sqrt{1}) + \frac{1}{4}\sum_{k=1}^5 \frac{1}{k^{3/2}} + \frac{1}{4}\sum_{k=6}^{99} \frac{1}{k^{3/2}} + \frac{1}{4}\sum_{k=100}^{999} \frac{1}{k^{3/2}} + \frac{1}{4}\sum_{k=1000}^{40000} \frac{1}{k^{3/2}} \\ &\le 2(\sqrt{40001} - \sqrt{1}) + \frac{1}{4}\sum_{k=1}^5 \frac{1}{k^{3/2}} + \frac{1}{4}\sum_{k=6}^{99} \frac{11}{k(k + 1)} + \frac{1}{4}\sum_{k=100}^{999} \frac{32}{k(k + 1)}\\ &\qquad + \frac{1}{4}\sum_{k=1000}^{40000} \frac{201}{k(k + 1)}\\ &= 2(\sqrt{40001} - \sqrt{1}) + \frac{1}{4}\sum_{k=1}^5 \frac{1}{k^{3/2}} + \frac{1}{4}\cdot (11/6 - 11/100) \\ &\qquad + \frac14\cdot (32/100 - 32/1000) + \frac14\cdot (201/1000 - 201/40001)\\ &= 2\sqrt{40001} - \frac{140034247}{120003000} + \frac{1}{16}\sqrt2 + \frac{1}{36}\sqrt3 + \frac{1}{100}\sqrt5\\ &< 399. \end{align*}

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Your sum is equal to

$$1 + \sum_{n=2}^{40000} \frac{1}{\sqrt{n}} < 1 + \int_1^{40000} \frac{1}{\sqrt{x}} \; dx = 1+ \left.2\sqrt{x}\right|_{x=1}^{40000} = 1+(2\sqrt{40000} - 2) = 399.$$

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  • $\begingroup$ Wouldn’t the sum be less than? $\endgroup$ Sep 24, 2021 at 13:29
  • $\begingroup$ @TymaGaidash Yes.....? $\endgroup$
    – B. Goddard
    Sep 24, 2021 at 13:29
  • $\begingroup$ You said “sum is equal to”. Right? Or maybe I understand something wrong. $\endgroup$ Sep 24, 2021 at 13:30
  • $\begingroup$ @TymaGaidash Yes. His sum is equal to the next thing I wrote. Then the very next symbol is "less than." $\endgroup$
    – B. Goddard
    Sep 24, 2021 at 13:31

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