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Please suggest a most simple sequence with the following properties:

$$\sum_{n=1}^{\infty} a_n=1$$

$$\frac1{a_n} \sim n!$$

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    I don't know what $\sim$ means here; could you edit the question to explain what you're looking for? It seems like the solution is trivial, because $\sum_{n=1}^\infty 1/(n!)$ converges.2011-04-10
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    It means $(1/a_n)/n! \to 1$ as $n \to \infty$.2011-04-10
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    At least it should mean that, according to standard notation. @Annix: What exactly $\sim$ means here?2011-04-10

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Let $a_n = 1/(n!)$ for $n \geq 2$. Then $\sum_{n=2}^\infty {a_n}$ converges to something, call the sum $L$. Let $a_1 = 1-L$. Then $\sum_{n=1}^\infty a_n = 1$.

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    $\sum_{n=0}^{\infty} \frac1{n!}$ will converge to e (it is exp(1)), and it is a trivial case to take $a_n=\frac1{e n!}$2011-04-10
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    @Anixx: And for your choice of $a_n$, how is it that $1/a_n \sim n!$ holds?2011-04-10
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    In this case $L=e-2$ and $a_1=3-e \approx 0.281718\ldots$2011-04-10
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Here's an example with all the $a_n$ rational.

$$ \sum_{n=1}^\infty \frac{n}{(n+1)!} = 1.$$

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    That's very nice. How did you find this sum?2011-04-10
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    @Douglas: Truncating the series expansion for $\exp(1)$. $\sum_{n=1}^\infty \frac{n}{(n+1)!} = \sum_{i=2}^\infty \frac{n-1}{n!} = \sum_{n=1}^\infty \frac{1}{n!} - \sum_{n=2}^\infty \frac{1}{n!}$.2011-04-10
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    @Douglas: It's many years ago now, but I first came across it as an undergraduate while answering some analysis questions (I remember thinking to myself, that's neat). It crops up from time to time on MSE, see for example http://math.stackexchange.com/questions/11665/rational-numbers-and-uniqueness/12982#129822011-04-11
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$a_n=1/(n!\sum_{n=1}^\infty 1/n!)$
(if $\sim$ means proportional)