# Write the Maclaurin series for function $f\left(x\right)=\frac{1}{3x+1}\:$

We have function $f:\mathbb{R}\rightarrow \mathbb{R}$ with $$f\left(x\right)=\frac{1}{3x+1}\:$$ $$x\in \left(-\frac{1}{3},\infty \right)$$ Write the Maclaurin series for this function.

Alright so from what I learned in class, the Maclaurin series is basically the Taylor series for when we have $x_o=0$ and we write the remainder in the Lagrange form. It has this shape: $f\left(x\right)=\left(T_n;of\right)\left(x\right)+\left(R^Ln;of\right)\left(x\right)=\sum _{k=0}^n\left(\frac{f^{\left(k\right)}\left(0\right)}{k!}x^n+\frac{f^{\left(n+1\right)}\left(c\right)}{\left(n+1\right)!}x^{n+1}\right)$

So when I compute derivatives of my function I can see that the form they take is:$$f^{\left(n\right)}\left(x\right)=\left(-1\right)^n\cdot \frac{3^n\cdot n!}{n!}x^n$$

Does that mean that the Maclaurin series is basically: $$f\left(x\right)=1-3x+3^2x^2-3^3x^3+....+\left(-1\right)^n\cdot 3^n\cdot x^n$$ ?

But what about that remainder in Lagrange form? I don't get that part. We didn't really have examples in class, so I've no idea if what I'm doing is correct. Can someone help me with this a bit?

Hint: Remember that

$$\frac{1}{1 + x} = \sum_{n=0}^{\infty} (-1)^n x^n \,\,\,\, \text{for} \,\,\,\ |x| <1$$

Then $$\frac{1}{1 +3x} = \ldots$$

for $|3x| < 1 \implies |x| < \frac{1}{3}$

• Wouldn't that just be $\frac{1}{1+3x}=\sum _{n=0}^{\infty }\left(-1\right)^n\cdot \:3^nx^n$ ? (What I wrote in my post) – MikhaelM Feb 4 '16 at 14:50
• That's right, you took the long way though. Compositions of functions works faster. – Aaron Maroja Feb 4 '16 at 14:52
• So is this all I need to do for this exercise? Is that my final MacLaurin series? – MikhaelM Feb 4 '16 at 14:53
• Yup, pretty much. – Aaron Maroja Feb 4 '16 at 14:53
• Oh, thank you! I thought it was harder than this. – MikhaelM Feb 4 '16 at 14:54

HINT: Use the fact that $\displaystyle \sum_\limits{n=0}^{\infty} x^n=\frac{1}{1-x}$ for $|x|<1$.