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I was wondering if: $$ \sum_{n=1}^{\infty} \pm \frac{1}{n} = 0 $$ have a predictable solution for $(\pm)$.

I have checked this equation in Python implementing the algorithm:

$$ s_1 = 1 $$ $$ s_{m+1} = s_{m} + \frac{1}{m},\quad if \quad |s_{m} + \frac{1}{m}| < |s_{m} - \frac{1}{m}| $$ $$ s_{m+1} = s_{m} - \frac{1}{m},\quad if \quad |s_{m} - \frac{1}{m}| < |s_{m} + \frac{1}{m}| $$

and this sum tends to $0$ very accurately as $\frac{1}{m}$.


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For this algorithm generated sequence of $+$ and $-$ seems to be random but it has some main rules, like there don't exist three times in a row the same operator: "$+++$" or "$---$".

Does there exist a function $f(m)$ which returns immediately $+$ and $-$ for any $m$ in correct order in respect to algorithm ?

Also I wonder if this equation have infinitely or finitely many solutions ?
Does this equation has a solution with pattern which could be easily demonstrated (I would love to see any) ?

Any words/articles/etc. that describes such considerations will be welcome, so I could google it.
I have no idea how to name such a problem.

Jean Marie
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Michal
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  • Sice the series is not abolutely convergent, there are infinitely many solutions: you can sum an arbitary, finite number of terms, and then substract until you get a negative number, etc. – ajotatxe May 01 '21 at 13:12
  • @ajotatxe's comment gives a countable infinity of series that sum to $0$, but there are even more than that: You can, for example, chose any signs you want for all the $3n+1$ terms, set the $3n+2$ terms to have the opposite signs, and then still reach $0$ by appropriate adjustments of the signs for the $3n$ terms. (This gives uncountably many solutions). – Troposphere May 01 '21 at 13:16
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    @Michal You may also write that $s_{m+1} = s_{m} - \frac{\text{sign}(s_{m})}{m}$. See also OEIS sequences https://oeis.org/A203811 , https://oeis.org/A203810 – Robert Z May 01 '21 at 13:19
  • Connected: https://math.stackexchange.com/q/742643 – Jean Marie May 01 '21 at 14:14
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    It's known that $\sum_{n=1}^\infty \frac{\mu(n)}{n} = 0$ – Jakobian May 01 '21 at 14:17
  • See as well this – Jean Marie May 01 '21 at 14:18
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    @Jakobian That's not an example since $\mu$ takes the value $0$ – jjagmath May 01 '21 at 14:18
  • @jjagmath yeah, sorry – Jakobian May 01 '21 at 14:19
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    If you treat signs $+1,-1$ as independent random variables (so called Rademacher functions) then the series converges for almost all choice of signs since variance is $\sum \frac{1}{n^2}$, i.e. it converges. The sum will be heavily concentrated around $0$ which can be deduced from estimates on partial sums, i.e. something like $P(|\sum| > t) < e^{-ct^2}$ for some constant c. – Salcio May 01 '21 at 14:23
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    How about this: Is there a periodic choice of signs so that $\sum \pm \frac{1}{n} = 0$ ? – GEdgar May 01 '21 at 14:38
  • As an extension of @GEdgar's question, how about this: For any $x\in [0,1]$ write $x$ as a binary fraction $(0.a_1a_2a_3)_2\ldots$ and set $f(x)=\sum_n (-1)^{a_n}\frac1n$ if the sum converges. GEdgar then asks whether the fiber $f^{-1}{0}$ interesects the rationals. We could also ask, e.g. does it intersect the algebraic reals? [Salcio essentially says that the domain of $f$ has measure $1$, and that $f$ is measurable and induces a nice probability measure on $\mathbb R$.] – Troposphere May 01 '21 at 21:27

2 Answers2

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Your solution can be characterized more simply by choosing the sign of $\frac 1n$ to be opposite the sign of $s_{n-1}$. Your solution does have three minus signs in a row. It starts $$\frac 11-\frac 12 - \frac 13 - \frac 14$$ That is the only time, as it is the only time the terms are decreasing so rapidly. If $s_n \gt 0$ and the sign on $\frac 1n$ is $+$, we must have had $s_{n-1} \lt 0$, so $s_n \lt \frac 1n$. Then the sign on $\frac 1{n+1}$ will be $-$. If $s_{n+1} \gt 0$ the sign on $\frac 1{n+2}$ will also be $-$, but as $\frac 1{n+1}+\frac 1{n+2} \gt \frac 1n$ we must cross $0$ this time and the sign on $\frac 1{n+3}$ will be $+$ so we will not get three $-$ signs in a row. The symmetric argument rules out three $+$ signs in a row.

Ross Millikan
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In general, since the series is not absolutely convergent, by suitably picking the signs you can obtain the result you desire. Here it is the proof.

Firstly over that for any $m$ the truncated sum

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

Since the whole series is divergent. This means that no matter which threshold $L$ we fix, the series will be greater at some point. In other words, there exist $k$ such that

$$ (**) \ \ \ \ \sum_{n=m}^{k} \frac{1}{n} > L $$ We will use this property repeatedly. Fix a limit $c \ge 0$ that you want to reach by properly adjusting the sign. Take the minimum $k_1$ such that

$$S_1 = \sum_{n=1}^{k_1} \frac{1}{n} \ge c$$

Analogously, there exist a minimum $k_2$ such that

$$ S_2 = S_1 + \sum_{n= k_1+1}^{k_2} - \frac{1}{n} \le c $$

By bouncing on and the off $c$ we get an $S_p$ that is closer and closer to it. Formally, note that for example since $k_1$ is the minimum with the "going above $c$ " property

$$ S_1 - \frac{1}{k_1} < c \ \ \Rightarrow \ \ \ |S_1 - c | \le \frac{1}{k_1} $$

The same argument shows that

$$| S_p - c | < \frac{1}{k_p}$$

So that $$\lim_{p \to \infty} S_p = c$$

As you can see, this is a fairly general proof: it holds for all non absolutely convergent series with infinitesimal term.

Andrea Marino
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  • I see that $\sum_{n=1}^{\infty} \pm f(n) = 0$ have infinitely solutions for all $f(n)$, if there exists at least one solution for $\sum_{n=1}^{\infty} \pm f(n) = \infty$ where $f(n + \epsilon) < f(n)$ and $\epsilon \gt 0$. Right ? – Michal May 02 '21 at 10:34
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    Yes! You don't need $f(n) $ to be decreasing, actually; the condition $\lim_{ n \to \infty } f(n) = 0$ (see the proofs) – Andrea Marino May 02 '21 at 19:02