How can I solve the following Note that I know if the sum starts from m=0, the term is equal to e^{x}. However, if m = n, where m>n and both are positive integers.
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Originally Posted by sjaffry How can I solve the following Note that I know if the sum starts from m=0, the term is equal to e^{x}. However, if m = n, where m>n and both are positive integers. Wolfram|Alpha: Computational Knowledge Engine Wolframalpha to the rescue!
How I would do it: $\displaystyle \sum_{m=n}^\infty \dfrac{x^m}{m!} = \sum_{m=0}^\infty \dfrac{x^m}{m!} - \sum_{m=0}^{n-1} \dfrac{x^m}{m!} = e^x-1-x-\dfrac{x^2}{2!} - \ldots - \dfrac{x^{n-1}}{(n-1)!}$
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