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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Sign up to join this communityImagine 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 $S_a>S_b$?
$$S_a=a_1-a_2+a_3-a_4+\cdots+a_n$$
$$S_b=b_1-b_2+b_3-b_4+\cdots+ b_n$$
$$S_b-S_a=\sum_{i=1}^{\lfloor{\frac{n}{2}}\rfloor} b_{2i-1}-a_{2i-1} + \sum_{i=1}^{\lfloor{\frac{n}{2}}\rfloor} a_{2i}-b_{2i} $$
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.
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.
No, say $a_1 = 1$, $b_1 = 0$, $a_2 = 10$, $b_2 = 1$, then $$ a_1 - a_2 = 1-10 = -9 $$ and $$ b_1 - b_2 = 0-1 = -1 $$