How do you find the maclaurin series expansion of #(e^x-1)/x#?

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How do you find the maclaurin series expansion of #(e^x-1)/x#?

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Answer 1 · 2015-10-06T15:00:41

#(e^x -1)/x= 1 + x/2 + x^2 /6 +x^3 /24 +...#

Explanation:

The maclaurin series expansion of #(e^x -1)/x# can be easily determined by using the maclaurin expansion of #e^x#.
So #e^x= 1+ x+x^2/(2!) +x^3/(3!) +x^4/(4!) +..........#

#e^x -1 =x+x^2/(2!) +x^3/(3!) +x^4/(4!) +..........#

#(e^x -1)/x= 1 + x/2 + x^2 /6 +x^3 /24 +...#

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How do you find the maclaurin series expansion of #(e^x-1)/x#?

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