Summation(n=0 to infinity) (lnx)²⁴/n! =1) x². 2)x. 3) lnx². 4)lnx

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Saurabh Koranglekar · Answer

Dear student To solve this problem Use expansion of e^t series summation wher t = ( ln x ) ^ 24 If not solved kindly tell Regards

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Summation(n=0 to infinity) (lnx)²⁴/n! =1) x². 2)x. 3) lnx². 4)lnx

Summation(n=0 to infinity) (lnx)²⁴/n! =1) x². 2)x. 3) lnx². 4)lnx

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Summation(n=0 to infinity) (lnx)²⁴/n! =1) x². 2)x. 3) lnx². 4)lnx

Summation(n=0 to infinity) (lnx)²⁴/n! =1) x². 2)x. 3) lnx². 4)lnx

1 Answers

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