Can You Solve A Cambridge Exam Question? Math Problem, 1802

What is the sum of the series 1/(1×2) + 1/(2×3) + 1/(3×4) + ... ? This problem was part of the Cambridge Exam in 1802. Can you figure it out? Watch the video for a solution.

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Cambridge problems. This problem is on page 22 of the document:

Check out Mathologer's video on Euler's series! It uses the series in this video to show the reciprocal sum of squares is bounded and therefore converges.

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[xtehmoonlight]

You mean they gave this "easy" question to make the candidate relaxed before hitting them with a comet question

[kolesov]

About the Moon and the Earth.
1. Consider an instantaneous change of Earth mass. Call this event the "Event". Assume that the Moon and the Earth can be treated as point objects for simplicity of calculations (you can take into account higher order corrections but I doubt that is what was needed for an exam 200 years ago). Also assume that the Earth is much heavier than the Moon, so we consider the Earth immobile instead of doing two-body problem exactly.
2. Before the Event, the Moon orbit was circular; from force balance we obtain v_0^2/r=(g*R^2/r^2), where v_0 is the velocity of the Moon, g is acceleration of free fall at the surface of the Earth, r is the distance between the Moon and the Earth, R is the radius of the Earth. Right after the Event, the velocity of the Moon is the same but g -> g/2, so there is no force balance any more. New g = a.
3. Calculate energy of the Moon after the Event. W = v_0^2/2-a*R^2/r = (g*R^2/r^2)*r-g*R^2/r = 0. It means that the orbit of the Moon is now parabolic. (If W<0 the orbit is elliptic, if W=0 orbit is parabolic, W>0 the orbit is hyperbolic). Moreover, since radial velocity of the Moon is zero, it means it is at the point of the closest approach to the Earth, and the Earth is in the focus of the orbit.
4. Now find the properties of the orbit. If initially the Earth is at (x=0, y=0) and the Moon is at (x=0, y=-r) then the equation of the trajectory is y = p*x*x-r, where p is unknown. Consider a point where y=0. Substitute it to the orbit equation: 0 = p*x*x-r, therefore, x = sqrt(r/p). Also, the derivative of the orbit y' = dy/dx gives us relation between v_y and v_x:
y' = v_y/v_x = 2*p*x.
Apply this equation at x = sqrt(r/p).
2*p*sqrt(r/p) = v_y/v_x. (1)
Make use of energy conservation: v^2 = v_x^2+v_y^2 = 2*a*R^2/sqrt(x*x+y*y)
Again, if x = sqrt(r/p), we obtain
v_x^2+v_y^2 = g*R^2/sqrt(r/p) (2).
Finally take into account angular momentum conservation
L = v_0*r = v_y*x = v_y*sqrt(r/p) (3)
Combine Eqs.(1-3) and initial expression for force balance v_0^2=g*R^2/r to get (if I am not mistaken in algebra) the following equation for p:
b^2+1/4-b=0,
where b=sqrt(pr). The only solution (and there MUST be only one solution!) b = 1/2, or p = 1/(4*r). Finally, the orbit equation reads as
y = x*x/(4*r) - r.
5. Discussion. Parabolic orbit is very unstable. Indeed, if the energy of the Moon is not exactly zero we would either have an elliptic or hyperbolic orbit. In this case it is important if the initial orbit of the Moon was exactly circular or not; if not, when exactly the Event happened; as the Event happened, what half of the Earth was removed (upper or lower with respect to the Moon direction), because it changes total energy; two body problem might be necessary to solve instead of simple model considered, etc.

Please like this solution if you agree with it.

[deidara_]

x=0
for i in range(1, 1000): x += 1/(i*(i+1)); print(x)

_very_ slowly approaches 1

[Lagadep]

I used power series, and I feel like I used a tank to kill a fly :')

[MindYourDecisions]

YouTube might be slightly broken? For some, the video preview is showing the solution instead of the problem which is really unfortunate. My guess is it's a bug related to the redesign as I never had this issue before (does anyone actually like the new design?). The good news: I told YouTube support about it and they are looking into the problem, thanks YouTube! I'm still seeing the wrong preview on Twitter/Facebook shares, but I am optimistic they will figure it out and my next video will post fine.

[twwc]

Your solution is very nice. I didn't think to use partial fractions. Instead, I evaluated the first few partial sums and saw that they were 1/2, 2/3, 3/4, 4/5, etc. I then used mathematical induction to prove the sum of the first n terms is n/(n+1), then took the limit as n goes to infinity to get 1.

[robinlindgren]

as soon as I saw the problem I immediately thought "this will be solved with a telescopic sum" but it was a long time since I did this type of problem and I forgot how to approach it beyond "I want to make a telescopic sum".

[adandap]

Here's another solution I found. Integrate the series S(x) = x + x^2/2 + x^3/3 +... term by term from 0 to 1 to get the series in question. But S(x) is just the Taylor's series for - ln(1-x), so our answer is (-1) x the integral from 0 to 1 of ln(1-x), which gives us 1.

[nikhilprasad]

I was confused too but as soon as you said "Partial Fraction", the solution was too straight forward and crystal clear.

[GordonHugenay]

lol, I've heard the term telescoping sum before, but never realized where it came from

[legendhero-eulc]

Thank you for the video! All of you friends are super awesome!

[tudortarta]

We get these kind of things at the math county olympiad in Romania....and it’s a miracle when you get these, they are like the easiest thing you could get

[JohnSmith-enyb]

The next question about the Moon orbit is interesting. If you calulate the orbital velocity v of the moon with the earths mass prior to impact -> v^2=GM/R (thats the equation for a circular orbit) You get something like a kilometer per second. But if the mass of the Earth M were to decrease by 50%, we'd take a look at another equation. Escape velocity. Because the two are related. The escape velocity of anything in orbit is exactly the square root of 2 times the circular orbit velocity. v^2=2GM/R If we then halve M, the 2 and the 1/2 cancel out. which means, that the moon would be at exactly escape velocity!!

[AudiMcAvoy]

I played with Steam's "Universe Sandbox^2" program. They actually have an Earth/Moon orbit simulation which allows you to instantaneously change Earth's mass to 0.5 Earths.

I did this several times, and it seems that the moon's position at the time of change makes a big difference in the results. Most of the time, I seen the orbit become very eccentric (from the original 0.06 to around 0.90). Most of the time the distance at closes approach was only minimally affected, but the distance at furthest approach was tripled, or more. The orbital period was typically in the 3 to 8 year range. However, on one occasion, the moon did reach escape velocity.

Pretty fun to play with. I highly recommend giving it a try yourself.

[johnbolin]

Honestly, I had it figured out before seeing your solution, without formula. Glad to know I still got it!

[UzumakiNaruto-rwtb]

I used that partial fraction method from integration to break the 1/n(n+1) term into 1/n -1/(n+1)

[isher__]

Calculus: how to write 1 and 0 in unnecessarily complex fashions.

[tharagleb]

I think it is easier to prove that the partial sum = k/(k+1) by induction, the limit as k goes to infinity is then 1.

[WhyDoesMyCodeNotCompile]

Thank you very much. I learned about partial fractions and the problem is a nice exercise

[AREmrys]

You can also use complete induction to prove that the kth partial sum of this series is k/(k+1) which is equal to 1 - 1/(k+1).