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Show that, for any positive integer \(n\geq 1\) it is true that: \[\displaystyle \sum_{k=1}^{n} \sqrt{k} \geq \sqrt{\displaystyle \sum_{k=1}^{n} k}.\]

Note by Hjalmar Orellana Soto
10 months, 3 weeks ago

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for \(n\geq 2\) it can be easily proven by applying triangle inequality recursively

assume a,b,c are three terms in the seuence

we have \(\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}\) as \(\sqrt{a},\sqrt{b},\sqrt{a+b}\) form sides of a right triangle

now we have \(\sqrt{a}+\sqrt{b}+\sqrt{c}\geq \sqrt{a+b}+\sqrt{c}\geq\sqrt{a+b+c}\) and so on....

sum of each of the outer sides of the spiral form the LHS of the equation,and RHS is given by the line to centre

also note that for n=1 the line to centre is same as side, the only point of equality

Anirudh Sreekumar - 10 months, 1 week ago

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very nice solution, Anirudh

Hjalmar Orellana Soto - 10 months, 1 week ago

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thank you :)

Anirudh Sreekumar - 10 months, 1 week ago

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\(LHS^2 = 1+2+\cdots+k + 2\sum_{1\le x<y\le n}\sqrt{xy}\ge 1+2+\cdots+n = RHS^2\) and equallity occurs when \(n=1\)

Reynan Henry - 10 months, 3 weeks ago

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In addition to the problem.... Is this necessary information for saying that the limit below converges? \[\lim_{x\to \infty} \dfrac{\displaystyle \sqrt{\sum_{n=1}^{x}n}}{\displaystyle \sum_{n=1}^{x}\sqrt{n}}\]

Hjalmar Orellana Soto - 10 months, 3 weeks ago

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(Assuming I'm interpreting you correctly,) the essence of what you're asking here is:

If \( a_n, b_n \) are sequences of (positive) reals, such that \( a_n \geq b_n \), does \( \lim \frac{ b_n } { a_n } \) exist?
If yes, why?
If no, what is a counter example? What other assumptions do we need to add to guarantee a true statement?

Calvin Lin Staff - 10 months, 3 weeks ago

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Exactly that is what I ask, I was thinking of asking for the value of \[\displaystyle \sum_{n=1}^{\infty} (\frac{\sqrt{\sum_{k=1}^{n}n}}{\sum_{k=1}^{n}\sqrt{n}})\] but I really don't know how to approach it

Hjalmar Orellana Soto - 10 months, 3 weeks ago

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@Hjalmar Orellana Soto The answer to "does \( \lim \frac{ b_n } { a_n } \) exist?" is no for general sequences. Do you see an obvious counter example?

For your recent comment, are you looking for the summation, or for the limit of the term (as in the prior comment)?

Calvin Lin Staff - 10 months, 3 weeks ago

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@Calvin Lin The answer to the limit question is yes for this case, using the same argument which is used for \(\lim_{x\to \infty} \frac{x}{e^x}\).... and I'm looking for the summation

Hjalmar Orellana Soto - 10 months, 3 weeks ago

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@Hjalmar Orellana Soto See How to ask for help to understand how to provide proper context of what you want. E.g. You can see that people quickly established the inequality, but that doesn't help with your concern.

Calvin Lin Staff - 10 months, 3 weeks ago

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@Calvin Lin The summation is something else, something I supposed and now I'm trying to approach, and I think the inequality may help for knowing if the summation converges... and then I'll try to approach the value of the summation, I don't know if I'm explaining right...

Hjalmar Orellana Soto - 10 months, 3 weeks ago

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Asymptotically, \(\sqrt{1} + \sqrt{2} + \dots + \sqrt{n}\) can be approximated by the integral \[\int_0^n \sqrt{x} \ dx = \frac{2}{3} n^{3/2}.\] And of course, \[\sqrt{1 + 2 + \dots + n} = \sqrt{\frac{n(n + 1)}{2}}.\] That should help answer your question.

Jon Haussmann - 10 months, 3 weeks ago

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We want to prove that \[\sqrt{1} + \sqrt{2} + \dots + \sqrt{n} \ge \sqrt{1 + 2 + \dots + n}.\] Squaring the left-hand side, we get a term of \(\sqrt{k} \cdot \sqrt{k} = k\) for each \(1 \le k \le n\). The inequality follows.

Jon Haussmann - 10 months, 3 weeks ago

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