Let's say you got a series $a_n$, such that $\sum^\infty_{n=0}a_n=L>0$. For any $0 \lt K \le L$, can you always find a subsequence $b_n$ of $a_n$, such that $\sum^\infty_{n=0}b_n=K$? If not always, when can you?
(Inspired by this comment.)
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Are we assuming that $\sum a_n$ converges? – hardmath Feb 09 '16 at 21:58
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1If $\sum_nx_n$ converges but $\sum_n|x_n|$ does not, then for any $r$ there is an infinite subsequence summing to $r$. – DanielWainfleet Feb 09 '16 at 22:42
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@user254665 Do you have a citation for that? That sounds cool! – Christopher King Feb 09 '16 at 22:43
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1I recall it is in the book Infinite Sequences And Series by Bromwich. Let $P={n :x_n>0}={f(n):n\in N}.; M={n:x_n<0}={g(n):n\in N}$ where $f,g$ are strictly increasing. Neither of $\sum_{n\in P}x_n, ;\sum_{n\in M}x_n}$ converges but $x_{f(n)}$ and $x_{g(n)}$ go to $0$ as $n\to \infty.$ Take a finite set $A_1$ of consecutive terms of $A={x_{f(n)}:n\in N}$ until their sum $S_1$ exceeds $r.$ Then take consecutive terms of $B={x_{g(n)}:n\in N}$ of higher subscript than those in $A_1$, until their sum $S_2$ satisfies $S_1+S_2<r.$ Then select from $A$ until.... – DanielWainfleet Feb 09 '16 at 23:06
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That's the (Levy-)Steinitz (derangement) theorem. Actually the first known formulation was by Grassmann some 50-60 years earlier, in the context of vector valued series. – Lutz Lehmann Feb 09 '16 at 23:06
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Now that I re-thought it, the statement in Bromwich was that we could re-arrange the whole series to get a sum of any value. But here we are asked for a subsequence. I'm not at all familiar with the history of it. – DanielWainfleet Feb 09 '16 at 23:10
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We cannot always do it. Look for example at the series $\sum_0^\infty \frac{1}{10^n}$.
André Nicolas
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For large $L$, take $a_0 = L - 1$ and $a_n = 2^{-n}$ for $n \geq 1$. Then for any subset $I\subset \mathbb{Z}^{\geq 0}$, the sum $s = \sum_{n\in I} a_n$ has $s \geq L -1$ if $0\in I$ and $s\leq 1$ otherwise.
anomaly
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