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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?

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  • $\begingroup$ Why do you think there's a relation between the two? $\endgroup$ – rubik Aug 2 '16 at 9:01
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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

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

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  • $\begingroup$ @MathIsTheWayOfLife You are welcome. $\endgroup$ – Olivier Oloa Aug 2 '16 at 9:45

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