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I came across a problem in "CRUX" :

Let 'n' be a positive integer. If one root of the quadratic equation $x^2 - ax + 2n = 0$ is equal to $ { \frac {1} {\sqrt {1}}} + { \frac {1} {\sqrt {2}}} + ... + { \frac {1} {\sqrt {n}}}$ , prove that $ 2{\sqrt{2n}} \leq {a} \leq 3{\sqrt{n}}$ .

I'm interested in the summation part.

WolframAplha gave the answer as $ {H_n}^{1/2}$ , where ${H_n}^{(r)}$ is the generalized harmonic number.

Any tips on how to tackle this (I have no idea about what is a harmonic number) ?

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    I doubt there is a nice expression for that, withou refering to generalised harmonic numbers. – 2017-01-11
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    What do you need this general formula for? There is no general fomula but there may be different ways to claculate this sum. – 2017-01-11
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    General formulas come in handy a lot, and if a good derivation is known, then it make life a whole lot simpler. – 2017-01-11
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    This does not seem to be true for any $a$. If $n=1$ then $1$ is supposed to be a root of $x^2 -ax+2$. But this is only true if $a=3$. And now if you consider $n=2$ then you will get contradiction with $a=3$. – 2017-01-11
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    The root of what exactly? Alternatively, what is the command you gave to WolframAlpha? – 2017-01-11
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    @lcv Root of the quadratic equation $x^2 - ax + 2n =0$. I gave the command to compute that nasty "reciprocal radical" suimmation. – 2017-01-11
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    You just need a good enough approximation to the sum. Using the first term in Euler-Maclaurin formula you get $H^{(1/2)}_n \approx 2 \sqrt{n} -2$ – 2017-01-11
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    @lcv which is weaker than my zeta limit thing :-) – 2017-01-11
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    @QUANTUM This is XY problem. Are you interested in the sum or in solving the problem? Because you don't need close formula for the sum to solve the issue. For example left inequality is quite trivial. It follows from the fact the the polynomial **has to have** roots (i.e. delta must be nonnegative). – 2017-01-11
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    Yes, I think it's a bit late to be saying you wanted to factor the quadratic with that sum over there... – 2017-01-11
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    You are asking for a very complicated thing, hoping to make a simple problem "simpler". – 2017-01-11
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    I'm not primarily interested in solving the problem. I just want to improve my "Mathematics GK". – 2017-01-11
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    :-P Btw, I improved it. – 2017-01-11
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    See another answer [**here**](http://math.stackexchange.com/questions/1727376/find-formula-for-frac1-sqrt-1-frac1-sqrt-2-cdots-frac1-sqrt-n/1727456#1727456) – 2017-01-11
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    Btw, what does GK mean? – 2017-01-13
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    @SimpleArt "General Knowledge" – 2017-01-13
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    Ah... Ok. Say, want to join a chat room? http://chat.stackexchange.com/rooms/51337/room-for-simple-art-and-thegreatduck – 2017-01-13

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Not a closed form, but we have some good approximations:

$$\sum_{k=1}^n\frac1{\sqrt k}\approx \zeta(1/2)+\sqrt n+\sqrt{n+1}$$

Where $\zeta(1/2)\approx-1.4603545$

Similarly,

$$\sum_{k=1}^n\frac1{\sqrt k}\approx \zeta(1/2)+2\sqrt{n+\frac12}$$

Here is a table of values:

$$\begin{array}{c|c|c|c}n&\sum_{k=1}^n\frac1{\sqrt k}&\sqrt n+\sqrt{n+1}&2\sqrt{n+\frac12}\\\hline3&2.2844570504&2.2716962988&2.2813028780\\9&4.7047701338&4.7019231514&4.7040594942\\27&9.0278784160&9.0273005360&9.0277339729\\81&16.5951438914&16.5950306293&16.5951155765\end{array}$$

As $n\to\infty$, it approaches the sum in question. You can see how well this tends to approximate.

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    That really helped a lot !!!! +1. – 2017-01-12
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    @QUANTUM :P I found a better one, though not much better (how much better should we ask for at this point?) – 2017-01-12
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    Just give the link...However, your previous answer suffices.. – 2017-01-12
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    @QUANTUM Did you want a [graph](https://www.desmos.com/calculator/dekgfuitxr)? The [Euler-Maclaurin summation formula](https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula)? Or you mean my updated answer? – 2017-01-12
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    Don't want to extend the chat... Thanks for your help. Those two links are more than sufficient... – 2017-01-12