This would be equivalent to sum of x^3 over x! too right? (Since you're just multiplying numerator and denominator by x)
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Replying to @InertialObservr
Right, what I'm saying is if you multiply the top and bottom by (n+1), it becomes (n+1)^3/(n+1)! [or more simply, n^3 / n!]
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Replying to @chrissmith_tree @InertialObservr
Yeah you’re right, and quite frankly sum(n^3/n!) looks nicer and really blew my mind.
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Replying to @zach_mescudi @chrissmith_tree
ah, nice i'll go ahead and post that
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@chrissmith_tree do you want me to tag you in a reply?1 reply 0 retweets 0 likes -
Replying to @InertialObservr @zach_mescudi
All good - just glad to contribute and repay your dope content in some small way
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Replying to @chrissmith_tree @zach_mescudi
thanks my dude! really appreciated, and nice observation. You've got an eye for mathematical beauty
12:32 PM - 17 Jul 2019
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