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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3I 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
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It means $(1/a_n)/n! \to 1$ as $n \to \infty$. – Shai Covo Apr 10 '11 at 12:29
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1At least it should mean that, according to standard notation. @Annix: What exactly $\sim$ means here? – Shai Covo Apr 10 '11 at 12:44
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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 $\sum_{n=1}^\infty a_n = 1$.
Carl Mummert
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1$\sum_{n=0}^{\infty} \frac1{n!}$ 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
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2@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
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Here's an example with all the $a_n$ rational.
$$ \sum_{n=1}^\infty \frac{n}{(n+1)!} = 1.$$
Derek Jennings
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@Douglas: Truncating the series expansion for $\exp(1)$. $\sum_{n=1}^\infty \frac{n}{(n+1)!} = \sum_{i=2}^\infty \frac{n-1}{n!} = \sum_{n=1}^\infty \frac{1}{n!} - \sum_{n=2}^\infty \frac{1}{n!}$. – cardinal Apr 10 '11 at 21:59
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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#12982 – Derek Jennings Apr 11 '11 at 07:43
