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May 18, 2018 at 20:43 vote accept CommunityBot
Aug 15, 2017 at 9:25 comment added Wadim Zudilin I have already given a solution in the yesterday comment: take, e.g., $y_{2n-1}=0$ (that is, $a_{2n}=a_{2n-1}$) and $y_{2n}=(-1)^n/\sqrt{n}$ (that is, $(a_{2n+1}/a_{2n})^2=1+(-1)^n/\sqrt{n}$). Then $\sum_{n=1}^\infty y_n^k$ converges for any $k\ge1$ different from 2, in particular, for $k=1$ and 3, and diverges for $k=2$. Thus $\sum_{n=1}^\infty x_n$ diverges. The connection between $x_n$ and $y_n$ (was that your question?) comes from solving $x(2+x)=y$.
Aug 15, 2017 at 2:09 answer added Terry Tao timeline score: 5
Aug 15, 2017 at 1:49 answer added Christian Remling timeline score: 3
S Aug 14, 2017 at 15:42 history suggested user113386 CC BY-SA 3.0
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S Aug 14, 2017 at 15:42
S Aug 14, 2017 at 15:08 history suggested user113386 CC BY-SA 3.0
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Aug 14, 2017 at 13:57 history edited user113386 CC BY-SA 3.0
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Aug 14, 2017 at 13:56 comment added user113386 @WadimZudilin: Cannot fetch last $y_n^2$. Where did 2 come from?
Aug 14, 2017 at 13:46 comment added Wadim Zudilin The summation over $n\ge1$ can be shifted to $n\ge n_0$ without changing the problem; in particular, $a_n>0$ for all $n$ can be dropped as $a_{n+1}/a_n\to1$ implies that $a_n$ is eventually sign-definite. Denote $x_n=(-1)^n(a_{n+1}/a_n-1)$ and $y_n=(-1)^n((a_{n+1}/a_n)^2-1)$; we have $x_n\to0$ and $y_n\to0$ as (from $a_{n+1}/a_n\to1$), so that $x_n=y_n/2-(-1)^ny_n^2/8+O(y_n^3)$ as $n\to\infty$. Assuming $\sum_{n\ge n_0}y_n$ converges, the divergence of $\sum_{n\ge n_0}x_n$ will potentially require $\sum_{n\ge n_0}(-1)^ny_n^2$ to diverge.
Aug 14, 2017 at 12:14 history edited user113386 CC BY-SA 3.0
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Aug 14, 2017 at 11:03 review First posts
Aug 14, 2017 at 11:11
Aug 14, 2017 at 11:00 history asked user113386 CC BY-SA 3.0