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I was working on the infinite sum $$\sum_{x=1}^\infty \frac{1}{x(2x+1)}$$ and I used partial fractions to split up the fraction $$\frac{1}{x(2x+1)}=\frac{1}{x}-\frac{2}{2x+1}$$ and then I wrote out the sum in expanded form: $$1-\frac{2}{3}+\frac{1}{2}-\frac{2}{5}+\frac{1}{3}-\frac{2}{7}+...$$ and then rearranged it a bit: $$1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+...$$ $$2-(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-...)$$ and since the sum inside of the parentheses is just the alternating harmonic series, which sums to $\ln 2$, I got $$2-\ln 2$$ Which is wrong. What went wrong? I notice that, in general, this kind of thing happens when I try to evaluate telescoping sums in the form $$\sum_{x=1}^\infty f(x)-f(ax+b)$$ and I think something is happening when I rearrange it. Perhaps it has something to do the frequency of $f(ax+b)$ and that, when I spread it out to make it cancel out with other terms, I am "decreasing" how many of them there really are because I'm getting rid of the one to one correspondence between the $f(x)$ and $f(ax+b)$ terms?

I can't wrap my head around this. Please help!

Franklin Pezzuti Dyer
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    When you have a conditionally-convergent series you cannot freely re-arrange the terms. For any such series, there is some arrangement of terms that converges to any number. – Doug M Jun 09 '17 at 19:37
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    That is not a telescoping series. – Angina Seng Jun 09 '17 at 19:37
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    Rearranging is not a plain thing as it looks at the first glance :). This is known a long time ago; for example, the formal series $1 - 1 + 1 - 1 + \cdots$ can be "rearranged" as $(1-1) + (1-1) + \cdots$ or $1 + (-1+1) + (-1+1) \cdots$. The former one seems to converge to $0$ and the latter seems to converge to $1$. – Yes Jun 09 '17 at 19:42
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    $\sum_{n=1}^\infty \frac{1}{n(2n+1)}$ is an absolutely convergent series, so you can rearrange terms. That doesn't mean you can split the terms into partial fractions and split and rearrange those. – sharding4 Jun 09 '17 at 19:43
  • @DougM How can it be proven that you can rearrange them to converge to any number? – Franklin Pezzuti Dyer Jun 09 '17 at 19:43
  • (1) It's only a "telescoping sum" if it is of the form $\sum_{k} (f(k)-f(k+1))$. (2) You can't re-arrange series that are not absolutely convergent and expect the sum to remain the same. – Thomas Andrews Jun 09 '17 at 19:44
  • @Frpzzd "how can it be proven that it can be rearranged to any number" this is the Riemann Series Theorem. Proof included in the linked article. – JMoravitz Jun 09 '17 at 19:48
  • @JMoravitz Okay, thanks. – Franklin Pezzuti Dyer Jun 09 '17 at 19:49
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    "and then rearranged it a bit: " STOP! rearranging "a bit" is not allowed. – fleablood Jun 09 '17 at 19:49
  • Suppose $P_n$ is the partial sum of the first $n$ positive terms and $S_n$ is the partial sum of the first $n$ negative terms. I say that there is an arrangement that converges to $C$. Find $i$ such that is the smallest number in the sequence of $P$ such that $P_{i} > C,$ Next find $j$ such that $P_{i}+S_{j} <C$. etc. This will produce a sequence converging to $C.$ – Doug M Jun 09 '17 at 19:51
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    Basically you can not "infinitely borrow". If, say, you are moving the nk term to the n place you are infinitely borrowing. The terms are one to one but you are using k times as many as one term then another. Example 1 + 1/2 + 1/4 + 1/8 + can be rearranged (1 + 1/4) + 1/2 + (1/16 + 1/64) + 1/8 + ... But whereas in the first you are adding an equal number of odd and even terms the second you are adding tweic as many even terms as odd terms. The results will be entirely different. – fleablood Jun 09 '17 at 19:58
  • A "rule of thumb" is to play $\mathsf{first}$ with a finite sum $\left(,\sum_{k = 1}^{n}\ldots,\right)$. It avoids to go to Wikipedia$\ldots$ – Felix Marin Jun 09 '17 at 20:18

3 Answers3

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When you split the series in question via partial fractions you create an alternating series that happens to be conditionally convergent (it's not absolutely convergent), so rearranging 'a bit' is not allowed. What went wrong is exactly what you surmised: you've lost the correspondence between terms in the two 'halves' of your alternating series, and so you can achieve the wrong sum.

If you're careful with your partial sums, however, you can confirm the following: $$ S_n := \sum_{k=1}^n \frac1{k(2k+1)}=\sum_{k=1}^n2\left(\frac1{2k}-\frac1{2k+1}\right)=2\left(1-A_{2n+1}\right), $$ where $$A_n:=\sum_{k=1}^n\frac{(-1)^{k+1}}k$$ is the $n$th partial sum of the alternating harmonic series, and therefore $S_n$ converges to $2(1-\log 2)$.

grand_chat
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By the Riemann rearrangement theorem, you have to be very careful when permuting terms of a conditionally but not absolutely convergent series. In your case it is probably simpler to notice that

$$ S=\sum_{n\geq 1}\frac{1}{n(2n+1)} = 2\sum_{n\geq 1}\left(\frac{1}{2n}-\frac{1}{2n+1}\right)=2\sum_{m\geq 2}\frac{(-1)^m}{m} $$ hence $$ S = 2 \sum_{m\geq 2}(-1)^m \int_{0}^{1}x^{m-1}\,dx = 2\int_{0}^{1}\frac{x}{1+x}\,dx = \color{red}{2(1-\log 2)}.$$

Jack D'Aurizio
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This is one of the most astonishing things in mathematics: every conditionally convergent series (meaning: that converges but is not absolutely convergent, such as the alternate harmonic series you mentioned) can be conveniently rearranged to converge to any arbitrary real number, or just diverge.

This is the celebrated Riemann Series Theorem. You can search for a quick introduction to this theorem in Wikipedia, but if you want a proof of the theorem look for Fiktengolz's book on Fundamentals of Mathematical Analysis. I hope this could help you.

Pedro A
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kvicente
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