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I found this simple infinite series of summed terms: 0/1 + 1/2 - 2/3 + 3/4 - 4/5 + 5/6 - 6/7 ....etc

When I tried calculating a bunch of terms, weirdly enough, it seems that an even number of terms produces ln(2) and an odd number of terms produces -(1-ln(2)). (You can test that out in this codepen I made)

My question is: How does this sequence produce the two numbers? Is there a certain property in it that results in this? Can a known sequence be rearranged and simplified to produce this sequence, or maybe this sequence re-written in a different form?

Thanks in advance!

Edit #1: I found that taking the known series for ln(2), substracting it from 1 and inversing addition and substraction results in this same series, which should calculate -(1-ln(2)) by definition . My question remains: How does it produce ln(2) and how does it alternate between the two?

iiSkLz_
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    You're probably thinking your sum goes addition, subtraction, addition, subtraction, etc. But actually, you started with two positive terms and then went positive, negative, positive, negative etc. which is inconsistent. For a series with a more uniform description, you would do 0/1-1/2+2/3-3/4+.... After that, compare what happens with an even number of summands versus an odd number in the partial sums. – anon Sep 25 '21 at 16:17

1 Answers1

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Let$$a_n=\frac{(-1)^{n-1}n}{n+1}.$$Then your series is $\sum_{n=1}^\infty a_n$. On the other hand, you have$$\sum_{n=1}^\infty\frac{(-1)^{n-1}}n=\log(2).$$And it is easy to check that, if $n\in\Bbb N$,

  • $\displaystyle\sum_{k=1}^{2n}\frac{(-1)^{k-1}}k=\sum_{k=1}^{2n-1}a_k$;
  • $\displaystyle\sum_{k=1}^{2n+1}\frac{(-1)^{k-1}}k=1+\sum_{k=1}^{2n}a_k$.

Therefore,$$\lim_{n\to\infty}\sum_{k=1}^{2n-1}a_k=\log(2)\quad\text{and}\quad\lim_{n\to\infty}\sum_{k=1}^{2n}a_k=-1+\log(2).$$

José Carlos Santos
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