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).
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).