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Power series of $\frac{1}{1+\frac{1}{4x}}$

Now in an attempt to find this power series I used the known power series of:

$\frac{1}{1+u} = 1-u+u^2-u^3+...$

Knowing this I simply substituted $\frac{1}{4x}$ as $u$ and found:

$\frac{1}{1+\frac{1}{4x}} = 1-\frac{1}{4x}+\frac{1}{16x^2}+...$

However this is not correct, so my questions is when am I allowed to do substitutions when comparing to known power series? Is it only when it is in the form cx where c is any real coefficient?

Thank you so much for your help. Kind Regards,

user2250537
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    Try rewriting your function as $4x(1+4x)^{-1}$. – Vectornaut Apr 12 '16 at 15:15
  • Why not use the radius of convergence. If you want a power series for $\frac{1}{1+\frac{1}{4x}}$ (about $x=0$) valid for small $x$, the take $\frac{4x}{1+4x}$ and use the known power series of $\frac{1}{1+u}$, valid for $|u|<1$. The series you derived is fine for large $|x|$. – almagest Apr 12 '16 at 15:17
  • You have found the Laurent series (on $|x| > 1/4). – mrf Apr 12 '16 at 15:19
  • Your series can be very useful, but it is technically not a power series, since power series by definition have the shape $\sum_{n=0}^\infty a_n(x-c)^n$. – André Nicolas Apr 12 '16 at 15:33

1 Answers1

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If the expansion is around $x=0$, then we can write for $|x|<1/4$

$$\begin{align} \frac{1}{1+\frac{1}{4x}}&=\frac{4x}{1+4x}\\\\ &=4x\sum_{n=0}^\infty(-4x)^n\\\\ &=\sum_{n=0}^\infty(-1)^n (4x)^{n+1} \end{align}$$

Note that the series in the original post, namely $$\frac{1}{1+\frac{1}{4x}}=\sum_{n=0}^\infty \left(\frac{-1}{4x}\right)^n$$is applicable for $|x|>1/4$.

Mark Viola
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