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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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2  
It's not true for $n=0$ :-) –  Robin Chapman Dec 2 '10 at 7:48
    
Indeed.. But I got the solution now (thanks to Derek) :) –  Listing Dec 2 '10 at 12:35

2 Answers 2

up vote 6 down vote accepted

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? $\left( \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)} + \cdots \right)<\sum _{k=1}^{\infty } \frac{1}{(n+1)^k}=\frac{1}{n}$ –  Listing Dec 1 '10 at 21:50
    
@user3123 Yes, sorry for the late reply, though no doubt you've sorted the problem now. –  Derek Jennings Dec 2 '10 at 8:09

Estimate it using a geometric series.

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Which side of the term? –  Listing Dec 1 '10 at 21:38
1  
The left one =) –  Jens Dec 1 '10 at 21:43

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