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I'm thinking of approaching this with the ratio test,

$\frac{a_{n+1}}{a_n}=\frac{x}{n+1}$. I know that the limit of this ratio as n approaches infinity is zero, therefore the series must converge?

Chad
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    Yes. If the limit is $0$ then the radius of convergence is infinite so it converges for all $x$. – Winther Nov 27 '16 at 23:42
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    Yes, it converges for any $x\in\Bbb R$ provided that the ratio test is passed for any $x$ (indeed it converges for any $z\in\Bbb C$ too.) – Masacroso Nov 27 '16 at 23:44
  • What is there to converge? Do you mean the corresponding sequence (infinite series)? – Jacob Wakem Nov 27 '16 at 23:58

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Your solution is correct. The series $$1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$ is the well-known Taylor series of expansion of $e^x$ about $x = 0$.

Ethan Alwaise
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