siriusmart@lemmy.worldM to Daily Maths Challenges@lemmy.world · 2 months ago[2024/05/16] Infinite multiplication serieslemmy.worldimagemessage-square2fedilinkarrow-up12arrow-down10file-text
arrow-up12arrow-down1image[2024/05/16] Infinite multiplication serieslemmy.worldsiriusmart@lemmy.worldM to Daily Maths Challenges@lemmy.world · 2 months agomessage-square2fedilinkfile-text
minus-squarezkfcfbzr@lemmy.worldlinkfedilinkEnglisharrow-up1·edit-22 months ago solution The terms can be rewritten as: (2/1) * (3/2) * (4/3) * … * ((n+1)/n) * … Each numerator will cancel with the next denominator. In total everything cancels, so the answer is the empty product, 1. …Wait… Uhm, ignore that. Rather, consider the products we get when multiplying. We get: 2/1. 6/2. 24/6. Etc. That is, we have: Π (n = 1 to k) (n+1)/n = (k+1)! / k! = (k+1)k!/k! = k+1 k+1 clearly goes to infinity as k → ∞, so our product diverges to infinity.
solution
The terms can be rewritten as:
(2/1) * (3/2) * (4/3) * … * ((n+1)/n) * …
Each numerator will cancel with the next denominator. In total everything cancels, so the answer is the empty product, 1.
…Wait…
Uhm, ignore that. Rather, consider the products we get when multiplying. We get: 2/1. 6/2. 24/6. Etc. That is, we have:
Π (n = 1 to k) (n+1)/n = (k+1)! / k! = (k+1)k!/k! = k+1
k+1 clearly goes to infinity as k → ∞, so our product diverges to infinity.