Sum of the first 86 square roots

Without using a calculator or a computer, can you compute the sum of square roots of the first 86 natural numbers, rounded to the nearest integer? In other words, we are looking for $$\sqrt{1} + \sqrt{2} + \ldots + \sqrt{86}$$ rounded to the nearest integer.

• Nice puzzle. Simple but polished. Sep 5, 2021 at 3:47

The given sum is what we would get if we were to compute the definite

integral of the square root function from $$0.5$$ to $$86.5$$ using fixed unit width bins and a simple mid point scheme.

But we can also compute this directly:

The integral is $$\int_{0.5}^{86.5} x^{0.5}dx = \frac 23x^{\frac 32}\Big|_{0.5}^{86.5} = \frac 23(86.5^{\frac 32} - 0.5^{\frac 32})$$.

Now by amazing coincidence $$9.3$$ squared happens to evaluate to

$$86.49$$

which is good enough to pass for $$86.5$$. One more lucky coincidence: $$9.3$$ is easily multiplied by $$\frac 23$$. So we need to do $$6.2$$ times $$86.5$$ which yields

$$536.3$$

The $$.3$$ cancels roughly with the $$\frac 23 \times 0.5^{\frac 32}$$

leaving us with

$$536$$.

Why is this so accurate? I.e., why does the humble midpoint rule give such a good approximation of the integral here?

An intuitive reason is that the sqrt is concave. To see this observe first that we can replace the tops of the approximating rectangles of the midpoint rule with tangents without changing the area. That way it is obvious that the midpoint rule systematically overestimates the integral. The good news is that the error is bounded by 1/8 times the change in slope. As the slope changes monotonically starting at $$\sqrt{1/2}$$ and never falling below zero the total error is bounded by $$\sqrt 2 / 16$$

• Well done! This is the solution I was looking for. I am glad you noticed the trick with 93. I picked 86 deliberately for that to work. Sep 5, 2021 at 4:21
• Nice. I wonder if it's possible to find a geometrical solution too. Sep 5, 2021 at 17:28
• @EricDuminil the integral argument is in essence geometric. We are essentially comparing the area inside a parabola to that of a slightly larger object. The error estimate in particular, can be translated into stacking certain triangles. Sep 5, 2021 at 18:56
• @loopywalt: Yes. I meant something different, though, e.g. the spiral of Theodorus. Sep 5, 2021 at 20:03
• @EricDuminil Looks more like a way of constructing all the square roots. I don't see an obvious way of getting at their sum from this approach. Sep 5, 2021 at 20:16

I'll take a shot at it

535

Arrived at as follows

First

The perfect squares are easy. 1 through 9, and add up to 45.

Then

the in-between numbers. I'm going to assume that they average halfway between the larger square and smaller square.

So

Between 1 and 4 - 2 numbers x 1.5 = 3
Between 4 and 9 - 4 numbers x 2.5 = 10
Between 9 and 16 - 6 numbers x 3.5 = 21
Between 16 and 25 - 8 numbers x 4.5 = 36
Between 25 and 36 - 10 numbers x 5.5 = 55
Between 36 and 49 - 12 numbers x 6.5 = 78
Between 49 and 64 - 14 numbers x 7.5 = 105
Between 64 and 81 - 16 numbers x 8.5 = 136
Between 81 and 100 - 18 numbers x 9.5 = 171

But

the numbers only go to 86, so 5 numbers x 9.125 (a guestimate) = 45.625, not 171

So

my guess is the sum of my estimates plus the perfect squares or
534.625 or 535 rounded off.

• Great start! There is another, more accurate method. Sep 5, 2021 at 1:20
• The square roots are biased uniformly high. I suspect $86$ was selected to make this fail. Sep 5, 2021 at 2:44