# Expansion of $e^x$ - correct form

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.

• No. There is a unique expansion of $e^x$. The second one you list is it. – mjw Apr 27 at 17:55
• The first one is a geometric series and sums to $\frac{1}{1-\frac{1}{x}} = \frac{x}{x-1}$ when $|x|>1.$ – mjw Apr 27 at 17:56
• The second provide an upper bound for $e^x$ for $x<1$. It is not equal to $e^x$ unless $x=0$. – Mark Viola Apr 27 at 17:57
• The second series is exactly equal to $e^x$ for all real numbers $x$. – Greg Martin Apr 27 at 18:00

## 1 Answer

Note the first series has a ratio of terms of $$1/x$$, thus, assuming $$|1/x| < 1$$, $$\sum_{k=0}^\infty \frac1{x^k} = \sum_{k=0}^\infty (1/x)^k = \frac{1}{1-(1/x)} = \frac{x}{x-1} = 1 + \frac{1}{x-1} \ne e^x.$$

• 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? – adriano Apr 28 at 19:26
• @adriano yes, that expansion is incorrect – gt6989b Apr 28 at 19:52
• thank you for confirming – adriano Apr 28 at 19:57