How to find the power series

Question

I have this function: $$ \frac1{(x+2)} $$ and I need to find an equal power series in the range of $$-2<x<2$$ I tried to take the power series of $$\ln(2+x)$$ and to use its derivative but I didn't get the right answer.

Answer 1

You have the power series $$\frac{1}{x+1}=\frac{1}{1-(-x)}=1-x+x^{2}-x^{3}+...=\sum_{n=0}^{\infty}(-1)^{n}x^{n}$$

which is valid for $|x|<1$ and called the geometric series.

Thus we have $$\frac{1}{x+2}=\frac{1}{2}\frac{1}{1+\frac{x}{2}}=\frac{1}{2}\frac{1}{1-(-\frac{x}{2})}$$ $$=\frac{1}{2}\sum_{n=0}^{\infty}(-1)^{n}(\frac{x}{2})^{n}$$ for $-2<x<2$.

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You also know the power series of $$\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}$$

for $|x|<1$.

So differentiating both sides gives $$\frac{1}{1+x}=\sum_{n=1}^{\infty}(-1)^{n-1}x^{n-1}=\sum_{n=0}^{\infty}(-1)^nx^n$$

which is the same as above and you can use that $$\ln(2+x)=\ln(2(1+\frac{x}{2}))=\ln(2)+\ln(1+\frac{x}{2})$$ to reach the same conclusion.

Answer 2

Recall that for $|a|<1$

$$\sum_{k=0}^\infty a^k = \frac1{1-a}$$

and

$$\frac1{x+2}=\frac1 2 \frac 1 {1+\frac x 2}$$