Is the sequence $u_n=n\sum_{i=1}^\infty2^{-i}(1-2^{-i})^n$ convergent when $n\to \infty$?

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If $i\leq \log n$ then $(1-2^{-i})^n<e^{-\frac{n}{2^i}}$. Unfortunately, I do not know what to do next

If $i\leq \log n$ then $(1-2^{-i})^n<e^{-\frac{n}{2^i}}$ Unfortunately, I do not know what to do next — piteer, Aug 06 '14 at 12:51
Added the remark to the body of question, so it does not get closed... also, why type \longrightarrow, when \to is so much shorter and makes sense syntactically? — , Aug 06 '14 at 13:18

score 2 · Answer 1 · edited Apr 13 '17 at 12:19

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This sequence does not converge. You can find the argument in Tossing Coins Until All Show Heads, American Mathematical Monthly, Vol. 101, January 1994, p.78-80. There are more details in my answer here.

edited Apr 13 '17 at 12:19

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answered Aug 06 '14 at 13:09

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Nice, +1. For those without access to the full text (such as me at the moment), the sequence exhibits an oscillating behavior between two finite positive limits, right? – Did Aug 06 '14 at 14:00
@Did Yes that is correct. I've added a link to another MSE answer of mine with more details. – Aug 06 '14 at 16:30

score 2 · Answer 2 · edited Mar 09 '17 at 17:31

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Some numerics may help. According to a rough calculation of the sum (capping the summation at $i=10^4$), it seems that the sum does not diverge, but stabilizes close to

$$u_n \approx 1.44268$$

enter image description here

A closer look at the graph shows that it oscillates around this value

enter image description here

edited Mar 09 '17 at 17:31

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answered Aug 06 '14 at 14:08

Winther

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Nice (especially your choice to show two graphs...). – Did Aug 06 '14 at 14:17

Is the sequence $u_n=n\sum_{i=1}^\infty2^{-i}(1-2^{-i})^n$ convergent when $n\to \infty$?

Problem

Progress

2 Answers2