By the ratio test, I know that this series convernges: $\sum2^{(-2^{n})}$, in the limit $n$ goes to infinity. Probably to something close to $.8$ (if not equal to $.8$). The problem is, how do I demonstrate it? or to what does it really converge? Any hint is appreciated.
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Are you looking for $\sum_{n=0}^{\infty}2^{(-2^{n})}$ or $\lim_{n\to\infty}\sum_{k=0}^{n}2^{(-2^{n})}$ ? – user127.0.0.1 Jan 21 '14 at 20:08
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If the series converge, what's the difference between the two expressions? – Spock Jan 21 '14 at 20:11
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1@Spock In the first series the summatio is over $n$ and in the second its over $k$, so the second would simplify to $\lim n2^{(-2^n)}$ – user127.0.0.1 Jan 21 '14 at 20:15
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I think he means the first one, the second converges to $0$. – user88595 Jan 21 '14 at 20:45
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1http://math.stackexchange.com/questions/583472/sum-of-infinite-series-1-1-2-1-4-1-16-cdots – Dan Shved Jan 21 '14 at 20:48
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Yes, it is the same question as in http://math.stackexchange.com/questions/583472/sum-of-infinite-series-1-1-2-1-4-1-16-cdots. Thanks – ace7047 Jan 21 '14 at 21:50