ln(2) series bijective equivalence

Asked Oct 20 '20 at 09:15

Active Oct 20 '20 at 10:00

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I am trying to get a bijective proof of the equivalence of the following infinite series which are easy to evaluate as $\ln(2)$ using elementary calculus tools:

$$\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{n}}$$

and

$$\sum_{n=1}^{\infty}{\frac{1}{n2^n}}$$

I tried grouping terms in (1) as has lower absolute value decreasing speed and is alternating to get terms in (2) as it has faster decrease and is positive defined. But no success on it.

edited Oct 20 '20 at 10:00

Bernard

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asked Oct 20 '20 at 09:15

24th_moonshine

2

What's a bijective proof? – Oct 20 '20 at 09:21
A bijective proof consists on grouping terms in both series to form group sums in a way to get a bijection between them. For instance $$\sum{\frac{(-1)^{n+1}}{n}}$$ is equivalent to $$\sum{\frac{1}{2n(2n-1)}}$$ as second is obtained grouping terms in the first series 2 by 2 sequentially. – 24th_moonshine Oct 20 '20 at 09:35
Just to add that no "increasing gaps" (sum more and more distant terms in each group) are allowed in those groupings in order to ensure that no term is left alone with no group when passing into the infinite sum approach. – 24th_moonshine Oct 20 '20 at 09:45
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One cannot expect that such a “grouping equivalence” always exists. As an example, the series $1+0+0+\ldots$ and $1/2+1/4+1/8 + \ldots$ converge to the same value, but no (finite) groups have the same sum. – Martin R Oct 20 '20 at 09:55
Agree on that, but also can not suppose it does not exist because sometimes doesn't. – 24th_moonshine Oct 20 '20 at 10:13
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Of course proof of non existence of such a bijection will be also recognized as an anwser to this question. – 24th_moonshine Oct 20 '20 at 11:25
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Does this answer your question? Sum the series of $\sum\limits_{n=1}^{\infty} \frac{1}{2^n n} $ – Dietrich Burde Dec 11 '21 at 19:25
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@DietrichBurde How does the linked question answer this one? – Mike Earnest Dec 12 '21 at 16:49

ln(2) series bijective equivalence

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