The Radical Brain Teaser That Stumped Everyone, And Changed Math Forever
89K views · Sep 14, 2025 · Science & Technology
Comments · 132
@Bruno_Haible · 1 year ago
This is the first time I see a glimpse into Ramanujan's thinking. Thanks!
66
@leonardhatcher3272 · 1 year ago
That burning smell is my brain cell going poof.
27
@soyoltoi · 11 months ago
7 years ago in high school, this problem peaked my interest in mathematics. Now I’m studying algebraic topology and applying for graduate school.
9
@mloneusk8269 · 1 year ago
Ramanujan need to be on India's currency, not Gandhi.
25
@dufo4766 · 1 year ago
I am just humbled by people like Ramanujan, thanks for posting!
6
@mohitrawat5225 · 1 year ago
Once Presh became a motivational speaker. He said to people - <br>When we work separately we get less but when we work together we get more result just like when a&b are squared separately we only get a²&b² but when we square a+b we get a²&b² with an extra 2ab😂😂😂😂
25
@Ackerman0205 · 1 year ago
The man who knew infinity
152
@cloysterd · 1 year ago
Whenever Ramanujan shows up I know I'm about to see something crazy that I barely understand, which he was able to figure out by himself with no formal education. Truly astonishing.
1
@bakamitaiyt · 1 year ago (edited)
<a href="https://www.youtube.com/watch?v=bGPKA9tb2CU&t=1000">16:40</a> A relatively minor point. You end up with an indeterminate form (infinity)^0, which needs to be separately evaluated. It does approach 1, but a rigorous proof would let L be the limit of the indeterminate form and show that ln(L), i.e. the limit of ln(n+2) / 2^(n-1), equals zero. As this is yet another indeterminate form (infinity)/(infinity), something along the lines of the Stolz–Cesàro theorem ("discrete l'Hôpital") may be used. That is, <br><br>ln(L)<br>= lim ln(n+2) / 2^(n-1)<br>= lim ( ln(n+2) - ln(n+1) ) / ( 2^(n-1) - 2^(n-2) ) <== Stolz–Cesàro<br>= lim ln (n+2 / n+1) / 2^(n-2)<br>= 0.
21
@BubbaYoga · 1 year ago
Fantastic.
2
@BossDropbear · 1 year ago
Finally this form of construction is similar to the series <br>1/(1-x) for x<>1 (and easiest to think about re convergence for 0<x<1)<br>=1+x/(1-x)<br>=1+x+[x^2]/(1-x)<br>=1+x+x^2+ x^3+ .... +[x^n]/(1-x)<br>This construction needs to consider convergence and the Domain for which it works.<br>R's brain teaser in same category which is why the video is so quick to go to that point.
2
@Muaz-ibn-masud · 1 year ago
Man with Infinity sight
1
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