A geometric proof that for 0<r<1, (1-r) + (1-r)r + (1-r)r² + ... = 1, and therefore that 1 + r + r² + ... = 1/(1 - r)pic.twitter.com/r22dM7PoUM
-
Show this thread
-
Replying to @AndrewM_Webb
I can show you an essentially similar (pictorial) proof of 1 + 2r + 3r^2 + ... = 1/(1 - r)^2. It's even prettier (and of course points beyond this, suggesting essentially similar proofs of all the analogous series that sum to 1/(1 - r)^n, n = 3, 4, ...)
1 reply 1 retweet 4 likes -
-
Replying to @AndrewM_Webb
Here's a starting point, showing how things go for r = 1/2:pic.twitter.com/asbc1ROJM6
2 replies 1 retweet 6 likes -
Replying to @MathPrinceps
Oh wow, okay! I need to think about this. Thank you
1 reply 0 retweets 1 like
Lemme know if you need more hints. It's probably a good idea to start with the r = 1/n case, and then work your way gradually up to r = m/n. The r = 1/3 case is similar enough to the r = 1/2 case to be fairly straightforward, given the foregoing picture.
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.