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I have a series $$1+ x+\frac{x^2}{2}+\frac{x^3}{4}+\frac{x^4}{8}...=1+\sum_{n=1}^\infty \frac{x^n}{2^{n-1}}$$

that looks an awful lot like a Taylor series of some kind. If the denominator of the fraction in the summation were $n!$ instead of $2^{n-1}$ we would have the Taylor series of $e^x$, expanded around $x=0$. What Taylor series is this?

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Hint

Can you evaluate

$$\sum_{n=1}^\infty \frac{x^n}{2^{n-1}}=2\sum_{n=1}^\infty \frac{x^n}{2^n}=2\sum_{n=1}^\infty \left(\frac{x}{2}\right)^n?$$

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    $\begingroup$ A geometric series? $\endgroup$ – NeutronStar Mar 23 '15 at 23:17
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    $\begingroup$ Yes. So you can solve the question. $\endgroup$ – mfl Mar 23 '15 at 23:30
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Hint: Divide the entire series by $2$, and then add $\frac 12$. You'll get something nicer.

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