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Using Stirling's formula, how can we bound the absolute error of $$e^x-\sum_{n=0}^N \frac{x^n}{n!}$$ on the interval $|x|\leq R$ where $R\leq N/2e$

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Use the upper bound by geometric series $$ \sum_{n=N+1}^\infty\frac{|x|^n}{n!}\le\frac{|x|^{N+1}}{(N+1)!}\cdot\frac1{1-\frac{|x|}{N+2}} $$ or use the Taylor expansion remainder term $$ e^{\theta x}\cdot\frac{x^{N+1}}{(N+1)!} $$ for some $\theta\in (0,1)$.

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  • $\begingroup$ Thank you. How does this provide a bound? $\endgroup$ – user19289 Jan 17 '18 at 12:50
  • $\begingroup$ Insert the Stirling approximation for $(N+1)!$ and apply the given bounds or similar ones to $\left(\dfrac{|x|e}{N+1}\right)^{N+1}$. $\endgroup$ – LutzL Jan 17 '18 at 14:11

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