rjf_berger on Nostr: npub1tfxfz…qn4hu About intuition: Take the continuous variant as an upper and lower ...
npub1tfxfz24qp6zykjngnxvp26fhztlj8h8srrf327afhtyy28rnptwshqn4hu (npub1tfx…n4hu) About intuition: Take the continuous variant as an upper and lower bound (stopping the integration either at n or n+1): \[ \int_0^{n/(n+1)} x^3 dx = (n+1/0)^4/4\] while the lhs. of your expression is \((\sum_{k=1}^n k)^2=(\frac{n(n+1)}{2})^2\). Thus there must be \[ \frac{(n+1)^4}{4} > (\sum_{k=1}^n k)^2=\frac{n^2(n+1)^2}{4} > \frac{n^4}{4}\] which is pretty obvious.
Published at
2023-05-25 11:29:02Event JSON
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"content": "nostr:npub1tfxfz24qp6zykjngnxvp26fhztlj8h8srrf327afhtyy28rnptwshqn4hu About intuition: Take the continuous variant as an upper and lower bound (stopping the integration either at n or n+1): \\[ \\int_0^{n/(n+1)} x^3 dx = (n+1/0)^4/4\\] while the lhs. of your expression is \\((\\sum_{k=1}^n k)^2=(\\frac{n(n+1)}{2})^2\\). Thus there must be \\[ \\frac{(n+1)^4}{4} \u003e (\\sum_{k=1}^n k)^2=\\frac{n^2(n+1)^2}{4} \u003e \\frac{n^4}{4}\\] which is pretty obvious.",
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