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The question calls $u=\frac{a}{x}$ and $u$ is the variable.

So for Taylor Series, we express it in $f(x)=\sum^{\infty}_{k=0}\frac{f^k(0)}{k!}x^k$

However, one hint says all we need is geometric series so we don't need to take derivatives. What does that mean.

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HINT:

$$\frac1{x-a}=-\frac1a\cdot\frac1{1-\left(\frac{x}a\right)}=-\frac1a\sum_{n\ge 0}\left(\frac{x}a\right)^n\;,$$

and

$$x+a=a-(-x)=a\left(1-\left(-\frac{x}a\right)\right)\;.$$

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  • $\begingroup$ Thanks a lot, that refreshed my memory $\endgroup$
    – CoolKid
    Apr 3 '16 at 19:46
  • $\begingroup$ @CoolKid: You’re welcome. $\endgroup$ Apr 3 '16 at 19:46
  • $\begingroup$ But since $u=\frac{a}{x}$, so $a=ux$. Why do you put $\frac{1}{a}$ outside the summation? $\endgroup$
    – CoolKid
    Apr 3 '16 at 20:06
  • $\begingroup$ @CoolKid: Is $u=\frac{a}x$, or is $u=\frac{x}a$? $\endgroup$ Apr 3 '16 at 20:08
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    $\begingroup$ @CoolKid: So far as I can tell from what you’ve written $a$ is a constant. Whether your goal is a power series in $x$ or one in $u$, $a$ can be manipulated just as if it were explicitly $2$, or $\pi$, or any other constant. Whether you take the sum and then multiply it by $\frac1a$, or multiply each term by $\frac1a$ and then take the sum, you get the same result. $\endgroup$ Apr 3 '16 at 20:16

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