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I have come across in a textbook to an expansion of e to the x in the following form: $$ 1+ \frac1x + \frac1{x^2} + \frac1{x^3} + \ldots $$ Is the above correct or is it a typo?

I am familiar with this type of expansion for e to the x: $$ 1+ \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots $$

Are they both correct? If yes, how is the first one above arrived at? I could not find it online anywhere.

Thank you in advance.

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    $\begingroup$ No. There is a unique expansion of $e^x$. The second one you list is it. $\endgroup$ – mjw Apr 27 at 17:55
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    $\begingroup$ The first one is a geometric series and sums to $\frac{1}{1-\frac{1}{x}} = \frac{x}{x-1}$ when $|x|>1.$ $\endgroup$ – mjw Apr 27 at 17:56
  • $\begingroup$ The second provide an upper bound for $e^x$ for $x<1$. It is not equal to $e^x$ unless $x=0$. $\endgroup$ – Mark Viola Apr 27 at 17:57
  • $\begingroup$ The second series is exactly equal to $e^x$ for all real numbers $x$. $\endgroup$ – Greg Martin Apr 27 at 18:00
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Note the first series has a ratio of terms of $1/x$, thus, assuming $|1/x| < 1$, $$ \sum_{k=0}^\infty \frac1{x^k} = \sum_{k=0}^\infty (1/x)^k = \frac{1}{1-(1/x)} = \frac{x}{x-1} = 1 + \frac{1}{x-1} \ne e^x. $$

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  • $\begingroup$ ok. So my math is not so great. but i guess you are saying that the textbook with that form of expansion for e^x is actually wrong then? $\endgroup$ – adriano Apr 28 at 19:26
  • $\begingroup$ @adriano yes, that expansion is incorrect $\endgroup$ – gt6989b Apr 28 at 19:52
  • $\begingroup$ thank you for confirming $\endgroup$ – adriano Apr 28 at 19:57

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