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

I am clueless here, I tried to multiply both sides with $n!$, but it doesn't make things better. I know that the left one converges against $e$ for $n=0$, but I better don't want to use its numerical value.

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    It's not true for $n=0$ :-)2010-12-02
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    Indeed.. But I got the solution now (thanks to Derek) :)2010-12-02

2 Answers 2

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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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    Do you mean this? $\left( \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)} + \cdots \right)<\sum _{k=1}^{\infty } \frac{1}{(n+1)^k}=\frac{1}{n}$2010-12-01
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    @user3123 Yes, sorry for the late reply, though no doubt you've sorted the problem now.2010-12-02
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Estimate it using a geometric series.

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    Which side of the term?2010-12-01
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    The left one =)2010-12-01