How to find the power series representation of [x - ln(x+1)]/x
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How to find the power series representation of [x - ln(x+1)]/x

[From: ] [author: ] [Date: 12-04-30] [Hit: ]
.= ∑[n ≥ 1] (-1)^{n + 1} x^n/(n + 1).......
find power series representation of [x-ln(x+1)] /x by starting with the formula 1+x+x^2 +x^3 +···=1/(1-x)

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Expressing ln(x + 1) as an integral, we obtain

ln(x + 1)

= ∫[y = t to x] dt/(1 + t)

= ∫[t = 0 to x] (1 - t + t^2 - t^3 +...) dt

= x - 1/2 x^2 + 1/3 x^3 - 1/4 x^ 4 + ....

Therefore

x - ln(x + 1) = 1/2 x^2 - 1/3 x^3 + 1/4 x^4 - 1/5 x^5 +...,

[x - ln(x + 1)]/x = 1/2 x - 1/3 x^2 + 1/4 x^3 - 1/5 x^4 +...

= ∑[n ≥ 1] (-1)^{n + 1} x^n/(n + 1).
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