Geometry of the Alternating Harmonic Series (visual proof)

This is a short, animated visual proof computing the sum of a the alternating harmonic series using a geometric argument. In particular, we recognize the sum of the alternating harmonic series as the area under the rectangular hyperbola, y=1/x over the interval [1,2]. To do this, we carefully describe a dissection process where we cut rectangles in half and replace half strategically with new rectangles, thus resulting in the final desired infinite series. We then use the fundamental theorem of calculus to compute the sum of the alternating harmonic series to be ln(2).

#math​ #infiniteseries #mtbos​ #manim​ #animation​ #theorem​ #pww​ #proofwithoutwords​ #visualproof​ #proof​ #iteachmath #calculus #harmonicseries #alternatingharmonicseries #series​ #mathvideo​ #naturallogarithm #logarithm #ln2 #Riemannsums #dissectionproof #dissection #area #integral #integralcalculus

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Music in this video:

[pyropulseIXXI]

If you arrange the series into blocks of 1 positive term followed by 36 negative terms, you will get a series that converges to exactly -ln 3

[RSLT]

This video is absolutely fascinating! I love how it presents a visual proof for computing the sum of the alternating harmonic series using a geometric argument. The dissection process and strategic placement of rectangles to achieve the desired infinite series is brilliant. It's amazing to see how the sum of the series can be recognized as the area under the rectangular hyperbola. The animation is top-notch and makes the concept very clear and engaging. Thank you for sharing this incredible visual proof and providing the link to the original source by Matt Hudelson. The music choice is also soothing and complements the video well. Keep up the great work, and I'll definitely consider subscribing and supporting your channel. Cheers! 🌟📐🔢

[apollo]

Amazing! I had never seen this proof and I can already say that it is my favorite way of showing the result of this sum

[Ninja]

That’s pretty cool. I knew that a geometric proof would involve the integral but i didnt know how to show that the sum actually gives the area under the curve.

The only way I’ve seen to prove this result before this was to use the maclaurin series of ln(1+x) and letting x=1.

[elevolucionestoica]

That was an outstanding class, truly impressive

[cheeseburgermonkey]

It only took 4 bars for me to realize this was going to turn into an integral

[ValkyRiver]

I interpreted the 1/1 9/8 5/4 11/8 3/2 13/8 7/4 15/8 2/1 as a musical scale C D E– Fǂ G Ad Bb↓ B– C

[morgangraley]

Elegant, intelligent, beautiful. Bravo! I had never conceptually understood this result; thank you for making it understood!

[andrewdjang]

Love this proof! I’ve seen it in pictorial form, but didn’t really get it until you walked me through it with the animations.

What would also be cool is to see a proof that the area under the hyperbola gives logarithms that doesn’t rely on the FTC!

[Yokuyin]

Do note that the Riemann rearrangement theory tells us you can rearrange the terms and get a different limit, as long as the sum of all positive terms and all negative terms diverge.

In this case, the sum of all negative terms is the harmonic series times -1/2, which diverges. And the sum of all positive terms has terms that are one reciprocal larger than the next term, which is the negative terms sum, thus it must also diverge.

For example, if you rearrange the terms such that every positive term is followed by the next two negative terms (1 - 1/2 - 1/4 + 1/3 - 1/6 - 1/8 + ...) the sum is ln(2)/2.

[yusufdenli]

Ln(1+x) taylor series;
ln(1+x)= x- (x^2)/2 +(x^3)/3 - (x^4)/4 +...

x=1
ln(1+1)= 1 - 1/2 + 1/3 - 1/4 +...
= ln(2)

[mh_uvdt_]

Hey plz read my comment... I have a great qn
Suppose u have 1 yellow colour highlighter. U can produce only yellow colour. Means 1 colour is produced with 1 highlighter.
If u have yellow and blue highlighter then u can produce blue, yellow and (yellow + blue=) green colour. Means if if u have 2 highlighters then u can produce 3 colours.
Similarly if u have 3 colour highlighter then u can produce 7 colours
So, how many colour can u produce with x number of highlighter(of diff

[Dark_Souls_]

I’m some years after an engineering degree and want to get back into studying math. I always wanted to do more proofs. Do you have a preferred text book for me to review?