# Please suggest a sequence with the following properties

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. –  Carl Mummert Apr 10 '11 at 12:12
It means $(1/a_n)/n! \to 1$ as $n \to \infty$. –  Shai Covo Apr 10 '11 at 12:29
At least it should mean that, according to standard notation. @Annix: What exactly $\sim$ means here? –  Shai Covo Apr 10 '11 at 12:44

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 $a_n=\frac1{e n!}$ –  Anixx Apr 10 '11 at 14:37
@Anixx: And for your choice of $a_n$, how is it that $1/a_n \sim n!$ holds? –  cardinal Apr 10 '11 at 14:59
In this case $L=e-2$ and $a_1=3-e \approx 0.281718\ldots$ –  Henry Apr 10 '11 at 15:41

$a_n=1/(n!\sum_{n=1}^\infty 1/n!)$
(if $\sim$ means proportional)

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Here's an example with all the $a_n$ rational.
$$\sum_{n=1}^\infty \frac{n}{(n+1)!} = 1.$$
@Douglas: Truncating the series expansion for $\exp(1)$. $\sum_{n=1}^\infty \frac{n}{(n+1)!} = \sum_{i=2}^\infty \frac{n-1}{n!} = \sum_{n=1}^\infty \frac{1}{n!} - \sum_{n=2}^\infty \frac{1}{n!}$. –  cardinal Apr 10 '11 at 21:59