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In what conditions the series converges?

$$\sum_{n=1}^{\infty}(2-e^a)(2-e^{a/2})\cdots(2-e^{a/n}), \quad a>0$$

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What have you tried so far? –  A.E Jul 7 '13 at 16:06
    
@Orangutango to guess the answer by trials. I need a rigorous proof though. –  user85369 Jul 7 '13 at 16:11
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The series converges if and only if $a=\ln2$ or $a\gt1$. Good, now that we have a flat answer to a flat question, let me suggest that you add what you tried to solve this question, why this failed, and so on, and we will be able to add why the answer is what it is. Shall we? –  Did Jul 7 '13 at 16:11
    
@Did $n$th root test? This was my best ace. –  user85369 Jul 7 '13 at 16:30
    
Really? You realize you are telling us you are unable to make a distinction between the divergence of $\sum\limits_n\frac1{\sqrt{n}}$ and the convergence of $\sum\limits_n\frac1{n^3}$... –  Did Jul 7 '13 at 17:05
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1 Answer

The general term of the given series is $$u_n=\prod_{k=1}^n\left(2-e^{a/k}\right)$$ and we have $$\frac{|u_{n+1}|}{|u_n|}=|2-e^{a/(n+1)}|=1-\frac{a}{n+1}+ o(\frac{1}{n+1})\quad \text{for $n$ large enough}$$ and using Raabe's test we conclude that the series is convergent if $a>1$.

Remark Since for $a=\log 2<1$ the general term $u_n=0$ then the series is also convergent for this value.

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The remark might be better written at the start, to justify applying the test, since we need denominator to be non-zero. –  Calvin Lin Jul 7 '13 at 17:55
    
@CalvinLin Notice that $u_n$ doesn't vanish only on $a=\log 2$ but also also on other values and in fact I gave just a formal proof. Thanks for your comment. –  Sami Ben Romdhane Jul 7 '13 at 18:00
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