A Nice Infinite Sum With Factorials

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[@farhansadik5423]

I was extremely happy hearing the 2e or not 2e joke again. Epic vid!

[@شعرکوتاه-ع7ظ]

Mathematics is difficult with some concepts unless someone teaches well and is a good teacher

[@SyberMath]

Aren't I a good teacher? 😜😁

[@DARWINFERNANDORIANOJIMENEZ]

Excelente

[@SyberMath]

Thanks!

[@johnnorwood3366]

Very nice!

[@SyberMath]

Thank you! 😍

[@pwmiles56]

In a generating function approach you go S = 2^2 + 3^2 /2! + 4^2 /3! exp(x) = x^0/0! + x^1/1! + x^2/2! + 4^2 x^3 /3! + ... x exp(x) = x^1/0! + x^2/1! + x^3/2! + x^4 /3! + (x exp x)' = x^0/0! + 2x/1! + 3x^2/2! + 4x^3 /3! + ... x(x exp x)' = x^1/0! + 2x^2/1! + 3x^3/2! + 4x^4 /3! + ... (x(x exp x)')' = x^0/0! + 2^2x^1/1! + 3^2x^2/2! + 4^2x^3 /3! + ... x=1 => LHS = S + 1 LHS = (x(x exp(x)+exp(x)))' = (x^2 exp(x)+x exp(x))' = (2x + x^2 + x + 1) exp x x = 1, LHS = 5e, S = 5e - 1

[@giom2306]

Much better than the teacher. Well done 👍

[@brandonfiennies7594]

Damn bro your a genius

[@pwmiles56]

@@brandonfiennies7594 Not really, it's a very well-known technique.

[@brandonfiennies7594]

@pwmiles56 I have to still get to university /college next year. I'll be studying acturial science. You seem like you did engineering

[@pwmiles56]

@@brandonfiennies7594 Maths, at Cambridge UK. I did later work in engineering but I had to re-learn the maths for it. Good luck with your actuarial ambitions (really!)

[@fahrenheit2101]

Neat!

[@scottleung9587]

Nice!

[@johns.8246]

Now do it for the alternating sum.

[@rainerzufall42]

sum = (e - 1) + e + 3 * e = 5 e - 1. For each term (n = 0, 1, 2, ...): (n+2)^2 / (n+1)! = ((n+3)(n+1)+1)/(n+1)! = 1/(n+1)! + n/n! + 3/n! = 1/(n+1)! + [*] 1/(n-1)! + 3/n!, [*] this term is 0 for (n=0)! Thus the whole sum is (e - 1) + e + 3 e = 5 e - 1.

[@leif1075]

WHY CHANGE THE INDEX AT ALL FOR GODS SAKE..I DONT see why ANYONE WOULD EVER EVER do that nio matter how smart..why not just.cancel ine n in the numerator with rhe denominator so you have n/n-1! Then that is easily.solved by rewriting as n-1 + 1/(n-1!) Which is 1/(n-2!) + 1/(n-1!)..Surely thisbis how everyone did it??

[@farhansadik5423]

Always watch with volume bro. He explained that if you didn't change the indices, and kept it as it was you would've gotten sum of 1/(n-2)! From n=1 to infinity. So, the first term of the sequence would've been 1/(-1)!. As you already know negative numbers don't have a defined factorial unless you use a special gamma function.

[@GourangaPL]

i instantly put it into wolfram and when i saw the answer is 5e-1 i had to watch the video how can that be

[@SyberMath]

Good thinking!