Series inequality $\sum _{k=n}^{\infty } \frac{1}{k!}\leq \frac{2}{n!}$

Show that : $\displaystyle\sum _{k=n}^{\infty } \frac{1}{k!}\leq \frac{2}{n!}$

I am unaware below, I attempted to increase both sides with $n!$, yet it does not make points far better. I recognize that the left one merges versus $e$ for $n=0$, yet I much better do not intend to utilize its mathematical value.

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2019-12-02 02:48:55
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Answers: 2

Estimate it making use of a geometric collection.

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2019-12-03 04:21:55
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Hint : $$\frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots $$ $$= \frac{1}{n!} \left( \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)} + \cdots \right) $$ $$ < \frac{1}{n!} \times \text{some geometric series}$$

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2019-12-03 04:21:28
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