<s>[INST] Taylor series of $(1+3x) \cdot \ln(1+x)$ I have to find Taylor series of $(1+3x) \cdot \ln(1+x)$. I know Taylor series of $(1+3x) \cdot \ln(1+x)$ but I do not know hot to simplify.
Any help?
 [/INST] Alternatively, you can calculate derivatives:
$$y=(1+3x) \cdot \ln(1+x) \quad (=0)\\
y'=3\ln (1+x)+\frac{1+3x}{1+x}=3\ln (1+x)+3-\frac2{1+x}\quad (=1)\\
y''=\frac3{1+x}+\frac2{(1+x)^2} \quad (=5)\\
y'''=-\frac3{(1+x)^2}-\frac{2\cdot 2}{(1+x)^3} \quad (=-7)\\
\vdots \\
y^{(n)}=\frac{(-1)^{n}\cdot 3\cdot (n-2)!}{(1+x)^{n-1}}+\frac{(-1)^n\cdot 2\cdot (n-1)!}{(1+x)^n}=\\
\frac{(-1)^{n}(n-2)!(3(x+1)+2(n-1))}{(1+x)^n},n\ge 2 \qquad (=(-1)^{n}(n-2)!(1+2n))$$
Hence:
$$y=\sum_{n=0}^{\infty} \frac{y^{(n)}(0)}{n!}x^n=y(0)+y'(0)x+\sum_{n=2}^{\infty} \frac{y^{(n)}(0)}{n!}x^n=\\
x+\sum_{n=2}^{\infty} \frac{(-1)^n(n-2)!(1+2n)}{n!}x^n=x+\sum_{n=2}^{\infty} \frac{(-1)^n(1+2n)}{n(n-1)}x^n.$$
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