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If we have an infinite sequence of natural numbers $A$ where all the elements in $A$ are unique and $A_n - A_{n-1}$ increases as $n$ increases, prove that $$\sum_{n=1}^\infty \frac{1}{A_n}$$ converges.

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I think you need to assume that $A$ is increasing, otherwise a sequence like $1,\frac{1}{2},\frac{1}{4},\ldots$ would be a counterexample. – Zev Chonoles Jan 21 '13 at 19:16
It is increasing, the question states that $A_n - A_{n-1}$ increases as $n$ increases. – user1825464 Jan 21 '13 at 19:23
I do not think it is true. Take $A$ such that $A_n - A_{n-1} = \ln n$ for every $n \geq 2$. We have $A_n \sim n \ln n$ but $\sum_{n=2}^\infty \frac{1}{n\ln n} = +\infty$. – Siméon Jan 21 '13 at 19:23
@user1825464: That is not the same thing. Note that for my sequence $A_n-A_{n-1}$ looks like $$-\tfrac{1}{2},\;-\tfrac{1}{4},\;-\tfrac{1}{8},\ldots$$ which is increasing. – Zev Chonoles Jan 21 '13 at 19:25
Now it's true: show that $A_n \ge n(n-1)/2$. – Robert Israel Jan 21 '13 at 19:35
up vote 3 down vote accepted

If we let $A_n=(n+3)\log (n+3)$, then $A_n-A_{n-1}$ increases, but the series does not converge.

How can the problem be made to work? We could for example assume that the $A_k$ are positive integers. Let $A_1=a$, and let $A_2-A_1=b$. Then $A_3-A_2\ge b+1$, $A_4-A_3\ge b+2$, $A_5-A_4\ge b+3$, and so on.

Thus $A_2\ge a+b$, $A_3\ge a+b+1$, $A_4\ge a+b+3$, $A_5\ge a+b+6$, and so on. Our sequence then essentially decreases at least as fast as $\frac{2}{n(n-1)}$, which is plenty fast enough for convergence.

Edit: The question changed, and now specifies that the $A_k$ are natural numbers. The answer has not been changed, since it already dealt with that case.

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My mistake -- I forgot to mention the important fact that $A$ is a sequence of natural numbers. Changing it now. – user1825464 Jan 21 '13 at 19:29
I had already guessed that might be a condition, so had typed an addition to the answer. – André Nicolas Jan 21 '13 at 19:36

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