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?
 A: The Maclaurin series of the given function is the product of the Taylor series of $\ln(1+x)$ and $1+3x$, treating the former as if it were a regular polynomial:
$$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\dots$$
$$3x\ln(1+x)=3x^2-\frac{3x^3}2+\frac{3x^4}3-\frac{3x^5}4+\dots$$
$$(1+3x)\ln(1+x)=x+\left(3-\frac12\right)x^2-\left(\frac32-\frac13\right)x^3+\left(\frac33-\frac14\right)x^4-\dots$$
$$=x+\sum_{k=2}^\infty(-1)^k\left(\frac3{k-1}-\frac1k\right)x^k$$
$$=x+\sum_{k=2}^\infty(-1)^k\frac{2(k+1)}{k(k-1)}x^k$$
A: We have that
$$\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$
therefore
\begin{align}(1+3x)\ln(1+x)&=\ln(1+x)+3x\ln(1+x) \\&= \sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}~+~\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(3x)x^n}{n}\\&=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(3x+1)x^n}{n}\\&=(3x+1)x-\frac{(3x+1)x^2}{2}+\frac{(3x+1)x^3}{3}-\frac{(3x+1)x^4}{4}+\cdots \\&=
x + \big(3-\frac{1}{2}\big)x^2-\big(\frac{3}{2}-\frac{1}{3}\big)x^3+\big(\frac{3}{3}-\frac{1}{4}\big)x^4-\big(\frac{3}{4}-\frac{1}{5}\big)x^5+\dots \\&=x+\frac{5}{2}x - \frac{7}{6}x^3+\frac{3}{4}x^4 - \frac{11}{20}x^5 +\dots
\\&= x + \sum_{n=2}^{\infty}(-1)^n\frac{2n+1}{(n-1) n}x^n
\end{align}
A: Know that
$$\ln{\left(1+x\right)}=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots$$
Then,
$\quad\left(1+3x\right)\ln{\left(1+x\right)}\\=\ln{\left(1+x\right)}+3x\ln{\left(1+x\right)}\\=\left(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots\right)+3x\left(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots\right)\\=x+\dfrac{5}{2}x^2-\dfrac{7}{6}x^3+\cdots$
A: Taylor series is expansion of a function about a given point $a$ in powers of $(x-a)$ when no point is mentionrd it could be taken as 0 so it becomes Mclaurin series.
Then $$f(x)=(1+3x)\ln(1+x)=  \sum_{k=1}^{\infty} (1+3x) (-1)^{k-1} \frac{x^k}{k}=\sum_{k=1}^{\infty} (-1)^{k-1} \frac{x^k}{k} + \sum_{k=1}^{\infty}(-1)^{k-1} \frac{3x^{k+1}}{k}.$$ 
Introducing $p=k+1$ in the second series $$\Rightarrow f(x)=x+\sum_{k=2}^{\infty}(-1)^{k-1} \frac{x^k}{k}+\sum_{p=2}^{\infty}(-1)^p \frac{3}{p-1}x^p.$$
$k$ being dummy we can replace it by $p$ in the first sum and we can write
$$f(x)=x+\sum_{p=2}^{\infty}(-1)^p \left( -\frac{1}{p} + \frac{3}{p-1}\right) x^p.$$
A: 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:
$$y=\sum_{n=0}^{\infty} \frac{y^{(n)}(0)}{n!}x^n=y(0)+y'(0)x+\sum_{n=2}^{\infty} \frac{y^{(n)}(0)}{n!}x^n=\\
x+\sum_{n=2}^{\infty} \frac{(-1)^n(n-2)!(1+2n)}{n!}x^n=x+\sum_{n=2}^{\infty} \frac{(-1)^n(1+2n)}{n(n-1)}x^n.$$
