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Find the power series expansion of $\frac{\ln(x)}{x}$ around the point $x=1$.

I was able to do this using the Cauchy product, but I'd like another way. If $\frac{\ln(t)}{t}=\sum_na_n(t-1)^n$, then $(\ln(x))^2=\int_1^x\frac{\ln(t)}{t}=\sum_na_n\int_1^x(t-1)^n=\sum_n\frac{a_n}{n+1}(x-1)^{n+1}$, but now you need to know the power series of $(\ln(x))^2$ to compare coefficients, which suggests using the Cauchy product again.

And if you take derivatives instead, $\frac{1-\ln(t)}{t^2}=\sum_nna_n(t-1)^{n-1}$, so maybe this works if you know the power series of $1/t^2$.

Does anyone have any suggestions/alternative methods?

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Setting $t=x-1$, you can use the power series expansion of $\;\ln x=\ln(1+t)$. You obtain the power series expansion of $$\frac {\ln x}x=\frac{\ln(1+t)}{1+t},$$ performing the division of the power series expansion of the numerator by the denominator by increasing powers of $t$.

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