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.

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

(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?

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
–
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
–
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
–
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
–
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...

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.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

very nice solution, Anirudh

Log in to reply

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 \(n=1\)

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.

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log 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)?

Log in to reply

Log 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.

SeeLog in to reply

Log in to reply

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