Find the power series representation of ln(1+x)/x?

Let's start with power series for 1/(1+x), since that's easy.

1/(1+x) = 1 − x + x² − x³ + x⁴ − x⁵ + ...

Integrate both sides:

ln(1+x) = C + x − x²/2 + x³/3 − x⁴/4 + x⁵/5 − x⁶/6 + ...

When x = 0, ln(1+x) = ln(1) = 0 ----> C = 0

ln(1+x) = x − x²/2 + x³/3 − x⁴/4 + x⁵/5 − x⁶/6 + ...

ln(1+x)/x = 1 − x/2 + x²/3 − x³/4 + x⁴/5 − x⁵/6 + ...

" D B " had the correct concept but ln ( 1 + x) = Σ { k = 1,2,..} ( -1) ^(k+1) x^k / k ===> ln ( 1 + x ) / x = 1 - x / 2 + x² / 3 - ...

I think you expand ln(1+x) with a tayler series expansion and divide each term in the expansion by x

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

/x = 1 + x/2 + (x^2)/3 .....

if you assume x is small you neglect x^2 and onward

ln(1+x)/x = 1 + x/2 for small x