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What would be the floor value of $1 + 1/\sqrt{2} + 1/\sqrt{3} + ... + 1/\sqrt{1000,000} ?$

I've considered telescoping, bijection, etc. but can't crack it. So perhaps squeezing is the way to go.

imranfat
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Curious George
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    @PedroPaulo The series is finite! – Theo Bendit Nov 22 '21 at 03:12
  • The integral test can be used to "squeeze" the sum of a convergent infinite series. But your series does not have infinite terms. So by drawing rectangles (the areas of these rectangles represent the value of the terms), as well as the function $y=\frac{1}{\sqrt{x}}$ in a smart manner, you can use its integral to get a floor value – imranfat Nov 22 '21 at 03:32
  • Why this is voted down? I think it is a valid question – Curious George Nov 22 '21 at 03:49
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    Does https://math.stackexchange.com/questions/509840/use-induction-to-prove-that-1-frac-1-sqrt2-frac-1-sqrt3[/this] answer your question? – Yuri Ginsburg Nov 22 '21 at 03:55
  • @YuriGinsburg Thanks, that's close already and provides a good way of thinking. 2 * \sqrt {1000,000} = 2000 in this case. But the actual answer is 1998. I still need to figure out how to get that answer. – Curious George Nov 22 '21 at 04:09
  • probably best to use $a=10$ and $b = 10^6$ in what follows, calculate the sum up to $\frac{1}{\sqrt 9}$ by calculator, then estimate $$ \int_a^{b+1} ; \frac{1}{\sqrt x} ; dx ; < ; \sum_{j=a}^b ; \frac{1}{\sqrt n} ; < ; \int_{a-1}^b ; \frac{1}{\sqrt x} ; dx $$ – Will Jagy Nov 22 '21 at 04:22
  • if the result of using $a=10$ is not good enough, try $a=11$ next. The overall mechanism is this: if we have $f(x) > 0$ but $f'(x) < 0,$ then $$ \int_a^{b+1} ; f(x) ; dx ; < ; \sum_{j=a}^b ; f(j) ; < ; \int_{a-1}^b ; f(x) ; dx $$ – Will Jagy Nov 22 '21 at 04:29
  • Integral proposed by @WillJagy $\int_1^{1000000} ; \frac{1}{\sqrt x} ; dx ; = 1998 ;$ . Actual sum is $1998.5391454911987$. In the linked article they are not looking for the floor value. – Yuri Ginsburg Nov 22 '21 at 05:00
  • @YuriGinsburg right. $2 \sqrt N + \zeta \left( \frac{1}{2} \right) $ is very close to the sum up to $N,$ while $2 \sqrt {N + \frac{1}{2} } + \zeta \left( \frac{1}{2} \right) $ is very very close – Will Jagy Nov 22 '21 at 05:04
  • There are some beautiful answers at https://math.stackexchange.com/questions/420733/proving-that-frac1-sqrt1-frac1-sqrt2-cdots-frac1-sqrt/420747#420747. – Curious George Nov 22 '21 at 05:15
  • @ThePianist sure. I was giving you an elementary treatment, without significant trickery, that you could do yourself. And learn something. – Will Jagy Nov 22 '21 at 05:20

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