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The Riemann Series Theorem says that any conditionally convergent series can be rearranged into any sum. Most examples I've seen have involved rearranging the series to 1/2 ln(2).

I was wondering how you would go about rearranging the series to equal any number. In this case I am interesting in having the alternating harmonic series equal 2.

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  • $\begingroup$ Take your calculator. Sum positive and negative terms to get above and below $2$ and record your choices. You might not find a closed form, but you should start noticing that you will get pretty fast to $2$. $\endgroup$ – Pedro Tamaroff Apr 30 '13 at 1:35
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To be conditionally convergent, the positive terms and negative terms must each be divergent on their own. Also, the terms must approach 0 in absolute value. Hence, take enough positive terms to be above the target (2 in your case), then take enough negative terms to be below the target, and continue. Since they each diverge, it will always be possible to overshoot the target. Since the terms approach 0, the amount of overshoot will shrink as you keep doing this.

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  • $\begingroup$ will there be a formula or specific way to write this or just a long list of numbers adding and subtracting? $\endgroup$ – cswinson Apr 30 '13 at 1:37
  • $\begingroup$ This is a nonconstructive proof because the conditionally convergent series was not specified. If you pick a particular series, with some tedious calculations, one could find out exactly how many + terms and - terms one needs to do this, but why? $\endgroup$ – vadim123 Apr 30 '13 at 1:39
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    $\begingroup$ I said alternating harmonic series $\endgroup$ – cswinson Apr 30 '13 at 1:40
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This is not an answer, but an oversized comment.

The following python code implements the method above. The code spits out a $1$ when the next positive term to be taken should be positive, otherwise a $0$.

c=2; #play with different values of c
l,a,b,s=[],[],[],0
for i in range(1,200):
    a.append(1/(2*i));
    b.append(1/(2*i-1));
a.reverse();
b.reverse();
for n in range(1,200):
    if s>c:
        s=s-a.pop();
        l.append(0);
    else:
        s=s+b.pop();
        l.append(1);
print(l);

The pattern for $c=2$ starts with $8 \ 1\text{s}, 0, 13 \ 1\text{s},0$ and after that the number of consecutive $1$s between each $0$ follows the repeating pattern $14,13,14$. For $c=0$ the repeating bit is $1,0,0,0$. For $c=1$, the repeating bit consists of $1,1,0$ repeated five times followed by $1,0$. When $c=-1$, the repeating bit is $29 \ 1\text{s}, 0, 30 \ 1\text{s}, 0$

Note: I haven't got round to proving the above statements, let alone finding a general formula for integer $c$, so inform me if you find any. In particular, must the $0,1$ sequence eventaully become repeating for all $c \in \mathbb{Z}$?

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