Approximation of Sum of square roots of n

Question

I recently came across a puzzle which was asked in an interview.

The puzzle requires us to compute sum of square root of first 50 natural numbers without any aid of computer (in fact just in head!).

Is there any mathematical way to calculate this or is he trying to test the computational ability of the participant?

He was probably trying to sound smarter than he is, or maybe he meant to ask a different question, unless you really were expected to come up with this in your head for some reason. Is it really the sum of the square roots or is it the square root of the sum? — , Sep 02 '16 at 13:23
Hm. Yeah, I'm calling shenanigans (on the interviewer, not you). — , Sep 02 '16 at 13:36
Let us say that , if I take a paper and pen to figure it out, Is there any way to approximate apart from binomial theorem? — mea43, Sep 02 '16 at 13:40

score 1 · Answer 1 · answered Sep 02 '16 at 13:42

1

Personally, I think the best way to get an approximation fast is to say $\sum_{i=1}^{50} \sqrt{i} $ is pretty much $\int_0^{50}\sqrt{x}dx = \frac{2}{3}\sqrt{50}\times 50 \approx 7^+ \times 50 \times \frac{2}{3} \approx 351 \times \frac{2}{3} = 234 $

A little off...

However, the first question coming to my mind is : what kind of job was it?

It happens that interviewers ask this kind of questions to catch people off-guard, but this seems just absurd.

answered Sep 02 '16 at 13:42

Vincent

2,318

The interview is for quant position in some hedge fund – mea43 Sep 02 '16 at 13:45
It looks fun. I hope you get it. – Vincent Sep 02 '16 at 13:47
What could it mean? That you have $50$ customers, and each customer $i$ looses $\sqrt{i}$ million dollars in the fund, and then what is the sum? – Dietrich Burde Sep 02 '16 at 14:01

Claude Leibovici · Answer 2 · 2016-09-02T17:56:25.660

Vincent gave a simple and elegant solution.

There is another possibility if you know generalized harmonic numbers since $$S_n=\sum_{i=1}^n \sqrt i=H_n^{\left(-\frac{1}{2}\right)}$$ Assuming that $n$ is large, the asymptotics is given by $$S_n=\frac{2 n^{3/2}}{3}+\frac{\sqrt{n}}{2}+\zeta \left(-\frac{1}{2}\right)+O\left(\frac{1}{n^{3/2}}\right)$$ but $\zeta \left(-\frac{1}{2}\right)\approx -0.207886$ which is quite small. Neglecting it, then $$S_n\approx \frac{4n+3}6 \sqrt n$$ If $n=50$, $\sqrt {50}\approx 7$ which makes $S_{50}\approx \frac {1421}6\approx 237$ while the exact value would be $\approx 239$.

Approximation of Sum of square roots of n

2 Answers2