TITLE: Computing a sum with a Taylor series
QUESTION [5 upvotes]: So I have a problem with the function:
$$f(x)= \ln[(1+3x)^{(1+x)}]$$
I want to find its Taylor series (Maclaurin series, because it's centered at $x=0$), and use it to calculate this sum:
$$\displaystyle\sum_{n=1}^{\infty} \frac{1-2n}{3^n n(n+1)}$$
So, I know the Taylor series for $\ln(1+x)$ , but I don't know how can I apply it here, or if I even should use that known formula?

REPLY [5 votes]: From the Maclaurin series expansion, one has
$$
\ln (1+x)= -\sum_{n=1}^\infty\frac{(-1)^n}nx^n, \quad |x|<1, \tag1
$$ one deduces
$$
\begin{align}
\ln[(1+3x)^{(1+x)}]&=(1+x)\ln(1+3x)
\\\\&=-(1+x)\sum_{n=1}^\infty\frac{(-1)^n}n3^nx^n
\\\\&=-\sum_{n=1}^\infty\frac{(-1)^n}n3^nx^n-\sum_{n=1}^\infty\frac{(-1)^n}n3^nx^{n+1}
\\\\&=3x-\sum_{n=2}^\infty\frac{(-1)^n}n3^nx^n-\sum_{n=2}^\infty\frac{(-1)^{n-1}}{n-1}3^{n-1}x^{n}
\\\\&=3x+\sum_{n=2}^\infty\frac{(-1)^{n-1}(2n-3)}{n(n-1)}3^{n-1}x^n
\\\\&=3x+\sum_{n=1}^\infty\frac{(-1)^n(2n-1)}{n(n+1)}3^{n}x^{n+1}. \tag2
\end{align}
$$
By putting $x=-\dfrac19$ in $(2)$, one gets

$$
\sum_{n=1}^{\infty} \frac{1-2n}{3^n n(n+1)}=3-8\ln\frac 32. \tag3
$$