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Imagine you have two finite alternating series.

$$S_a=a_1-a_2+a_3-a_4+\cdots+a_n$$

$$S_b=b_1-b_2+b_3-b_4+\cdots+ b_n$$

Question: If $|a_i|>|b_i|$ is $S_a>S_b$?

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  • $\begingroup$ No. Choose $a_1 = 2, a_2 = 1, a_3, ..., a_n = \frac 1 2$, together with $b_1 = 1, b_2, ..., b_n = 0$. When $n$ has the right parity, $S_a = S_b$. $\endgroup$ – user61527 Apr 10 '14 at 18:15
  • $\begingroup$ What is the relation between $a_i$ and $b_i$? $\endgroup$ – Zeonatra Apr 10 '14 at 18:16
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$$S_a=a_1-a_2+a_3-a_4+\cdots+a_n$$

$$S_b=b_1-b_2+b_3-b_4+\cdots+ b_n$$

$$S_b-S_a=\sum_{i=1}^{\lfloor{\frac{n}{2}}\rfloor} b_{2i-1}-a_{2i-1} + \sum_{i=1}^{\lfloor{\frac{n}{2}}\rfloor} a_{2i}-b_{2i} $$

From that one clearly one can make the difference in the odd members small and in the even members great or visceversa to obtain any sign.

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No, suppose all $a_i$ are equal and set to a large positive number $N$, and $n$ is even. Then $S_a = 0$. Certainly you can define an alternating sequence of numbers with absolute value less than $N$ such that the sum $S_b$ is positive.

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No, say $a_1 = 1$, $b_1 = 0$, $a_2 = 10$, $b_2 = 1$, then $$ a_1 - a_2 = 1-10 = -9 $$ and $$ b_1 - b_2 = 0-1 = -1 $$

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  • $\begingroup$ Also each $a_i$ and $b_i$ is positive $\endgroup$ – user122523 Apr 10 '14 at 18:22
  • $\begingroup$ Changed it. Same conclusion. $\endgroup$ – Marc Apr 10 '14 at 18:33

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