How can I proceed if I want to find a $M$ (non negative) so that $\sum_{n=M+1}^\infty \frac{1}{n!}<10^{-8}$
Can I use this taylor formula??
Thanks in advance!
2026-04-08 18:02:26.1775671346
Taylor series problem!
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\begin{align} \sum_{n=M+1}^\infty \frac{1}{n!} &= \frac{1}{(M+1)!} \left(1 + \frac{1}{(M+2)} + \frac{1}{(M+2)(M+3)} + \dots \right) \\ &<\frac{1}{(M+1)!} \left(1 + \frac{1}{(M+2)} + \frac{1}{(M+2)^2} + \dots \right) \\ &\le\frac{2}{(M+1)!}, \quad \text{ if } M+2 \ge 2 \end{align} Since $\frac{2}{12!} < 10^{-8}$, $M=11$ will do the trick.