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Let $x$ be a positive integer, let $b_k=\dfrac{x+2^k-1}{2^k}$, and let $n\in\mathbb{N}$ be as large as possible such that $b_n$ is a positive integer. Consider the sequence $(b_k)_{k=1}^n$. For example, taking $x=17$, we retrieve the sequence $(9,5,3,2)$. Now consider the sum $S(n)=\sum_{k=1}^{n}\frac{1}{b_k}$. How does this behave as $n\rightarrow\infty$?

j0equ1nn
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  • Hi Martin, you are right. I put more precision on the question/case: The sequence stops at its lowest integer. In the given case (x=17) for example a $b_5,b_6,\ldots$ does not exist. Hope this helps and it may be much more easier to show that the series is convergent/limited. –  Mar 08 '20 at 08:39
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    I should ask: How we can show/argue that the sum/series is definitely limited. What is its limit? When playing with the product $(1+\frac{1}{b_1})(1+\frac{1}{b_2})\ldots$ I see it always stays below 4. –  Mar 08 '20 at 08:46
  • When the sum $(1/b_i)$ is limited, then the product $(1+1/b_i)$ should be limited as well. Thats why I hope it is possible to show the limit of that sum. –  Mar 08 '20 at 08:56

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If we factor $x-1=2^k y$ where $y$ is odd, then $b_n = 2^{k-n} y + 1$ for all $n$, and thus $b_k$ is the last integral element of the sequence. The sum in question is then $$ \sum_{i=1}^k \frac1{b_i} = \sum_{i=1}^k \frac1{2^{k-i} y + 1} < \sum_{i=1}^k \frac1{2^{k-i} y} = \frac1y\bigg( 2-\frac1{2^{k-1}} \bigg) < 2. $$ The actual best upper bound, when taking $x=2^k+1$ for larger and larger $k$, is approximately $1.2644997803484442092$.

Greg Martin
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  • Thanks for that great explanation! Can we derive from that finding the limit of the product $\prod_{i=1}^n(1+\frac{1}{b_i})$? Based on the fact that the sum is lower than 2, does the product therefore stays below 4? –  Mar 08 '20 at 10:25
  • Just checked with K. Knopp "Theorie und Anwendung der unendlichen Reihen": The upper bound of the product seems to be $e^{\sum\ldots}=e^{1.2644...}$ –  Mar 08 '20 at 10:35