Bounding the Basel Problem Series (visual proof)

In this short video, we show that the series of reciprocals of squares is bounded above by 2 using a visual argument. This implies that the series converges by the monotone sequence theorem. This visual essentially also shows the process of Cauchy condensation.

This animation is based on an article by Ethan Berkove:

Another Look at Some p-Series
The College Mathematics Journal, Vol. 37, No. 5 (Nov., 2006), pp. 385-386 (2 pages)

#manim #euler #calculus #series #infiniteseries #baselproblem #mathvideo​ #math​ #mtbos​ ​ #animation​​​ #iteachmath #mathematics

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[lockaltube]

sure, let's draw a circle with r = sqrt(pi) and fit all these squares there

[alejrandom]

I like how both the convergence of the Basel series as well as the divergence of the Harmonic series use inverse of powers of 2

[AIexPlayGames]

I really like this sum. A few weeks ago I stumbled upon a proof of the basel problem on my own as well. (Obviously just a variation of an already known proof, but I am happy with finding it independently)

[YellowBunny]

I was thinking about improving this bond by cutting out the top right "triangle". But the width of each row is the difference of two harmonic numbers, so I don't think the curve of that "triangle" can be described by a nice function. There might be some integral or differential equation to define that curve, but I'll leave further thoughts about this as an exercise for the reader.

[deleted-something]

This reminds me about how little we know about gamma, even though is pretty related with how these series work

[vikramrawat]

Next video:Basel problem by visual proof 😁

[diribigal]

I'm a little disappointed that the 1/4 row has the squares of 1/16 all left-aligned instead of aligned to start at the quarter-marks (and similarly for the rows above). If you had aligned them that way, then the idea for why they fit might be obvious from the picture instead of requiring the audio

[FoxfNight]

sum 1/n^2 = pi^2/6 could be thought of as showing that the lenght of the vector (1, 1/2, 1/3, 1/4, ...) is pi/sqrt(6). You'd need to find a way to use infinite dimensional geometry for that to work or find a work around.

Edit:
Here's a workaround:
Draw a line segment of length 1, then a perpendicular line of length 1/2 such that it forms a right triangle.
Now draw another line of length 1/3 perpendicular to the hypothenuse to form another right triangle.
Now draw another line of length 1/4 perpendicular to the new hypothenuse.
As you itterate this process you'll approach a hypothenuse of length pi/sqrt(6). Maybe you can try to form a square of area pi^2/6 from it?
I haven't yet found a way to geometrically show the result.

[pixelpower.random]

I love the proof on the channel 3 Blue 1 Brown

[yousefidris]

I am shocked how you were able to do this on manim

[FoxfNight]

I sent a comment of my idea and now I have a Desmos representation of it, but YT does not allow me to send links. Do you have a way for me to send it to you if you're interested?

[zushyart]

Have you made a visual proof of the pi^2/6 sum?

[NandanGaming]

Next video idea:-
Explain how calculus works and introduction of it