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

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TopNewestfor \(n\geq 2\) 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 · 7 months ago

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– Hjalmar Orellana Soto · 7 months ago

very nice solution, AnirudhLog in to reply

– Anirudh Sreekumar · 7 months ago

thank you :)Log in to reply

\(LHS^2 = 1+2+\cdots+k + 2\sum_{1\le x<y\le n}\sqrt{xy}\ge 1+2+\cdots+n = RHS^2\) and equallity occurs when \(n=1\) – Reynan Henry · 7 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 · 7 months, 3 weeks ago

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– Hjalmar Orellana Soto · 7 months, 2 weeks ago

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 itLog in to reply

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 · 7 months, 2 weeks ago

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– Hjalmar Orellana Soto · 7 months, 2 weeks ago

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 summationLog in to reply

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 · 7 months, 2 weeks ago

SeeLog in to reply

– Hjalmar Orellana Soto · 7 months, 2 weeks ago

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...Log in to reply

– Jon Haussmann · 7 months, 2 weeks ago

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.Log in to reply

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 · 7 months, 3 weeks ago

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