How to derive a general formula for the expression $\sum_{k=1}^{n} { \frac {1} {\sqrt{k} } } $

Question

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) ?

I doubt there is a nice expression for that, withou refering to generalised harmonic numbers. — Henrik supports the community, Jan 11 '17 at 15:11
What do you need this general formula for? There is no general fomula but there may be different ways to claculate this sum. — Math-fun, Jan 11 '17 at 15:11
General formulas come in handy a lot, and if a good derivation is known, then it make life a whole lot simpler. — , Jan 11 '17 at 15:13
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 $a=3$. — freakish, Jan 11 '17 at 15:15
The root of what exactly? Alternatively, what is the command you gave to WolframAlpha? — lcv, Jan 11 '17 at 15:18
@lcv Root of the quadratic equation $x^2 - ax + 2n =0$. I gave the command to compute that nasty "reciprocal radical" suimmation. — , Jan 11 '17 at 15:21
You just need a good enough approximation to the sum. Using the first term in Euler-Maclaurin formula you get $H^{(1/2)}_n \approx 2 \sqrt{n} -2$ — lcv, Jan 11 '17 at 15:28
@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). — freakish, Jan 11 '17 at 15:30
Yes, I think it's a bit late to be saying you wanted to factor the quadratic with that sum over there... — Simply Beautiful Art, Jan 11 '17 at 15:31
You are asking for a very complicated thing, hoping to make a simple problem "simpler". — N. S., Jan 11 '17 at 15:36
I'm not primarily interested in solving the problem. I just want to improve my "Mathematics GK". — , Jan 11 '17 at 15:40
Ah... Ok. Say, want to join a chat room? http://chat.stackexchange.com/rooms/51337/room-for-simple-art-and-thegreatduck — Simply Beautiful Art, Jan 13 '17 at 12:58

Simply Beautiful Art · Accepted Answer · 2017-01-12T01:39:22.760

4

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.

edited Jan 12 '17 at 01:39

answered Jan 11 '17 at 15:24

Simply Beautiful Art

74,685

That really helped a lot !!!! +1. – Jan 12 '17 at 00:45
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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?) – Simply Beautiful Art Jan 12 '17 at 01:39
Just give the link...However, your previous answer suffices.. – Jan 12 '17 at 01:41
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@QUANTUM Did you want a graph? The Euler-Maclaurin summation formula? Or you mean my updated answer? – Simply Beautiful Art Jan 12 '17 at 01:45
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Don't want to extend the chat... Thanks for your help. Those two links are more than sufficient... – Jan 12 '17 at 01:47

How to derive a general formula for the expression $\sum_{k=1}^{n} { \frac {1} {\sqrt{k} } } $

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