Summation (n=0 to infinity) (logx)^24/n! = 1)x² 2) x. 3) log(1+x). 4) logx²

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Saurabh Koranglekar · Answer

Dear student Use e^t expansion in the question t = ( logx) ^ 24 Regards

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Summation (n=0 to infinity) (logx)^24/n! = 1)x² 2) x. 3) log(1+x). 4) logx²

Summation (n=0 to infinity) (logx)^24/n! = 1)x² 2) x. 3) log(1+x). 4) logx²

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Summation (n=0 to infinity) (logx)^24/n! = 1)x² 2) x. 3) log(1+x). 4) logx²

Summation (n=0 to infinity) (logx)^24/n! = 1)x² 2) x. 3) log(1+x). 4) logx²

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