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Is there a simple, elementary proof of the fact that: $$\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)=0$$ I have thought of a very simple notation for "harmonic" sums like these: just write down the numerators. So, for example:
$[\overline{1}]=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots=\infty\;$ is the harmonic series
$[\overline{1,-1}]=\frac{1}{1}+\frac{-1}{2}+\frac{1}{3}+\dots=\ln2\;$ is well known
$[\overline{1,1,-2}]=\frac{1}{1}+\frac{1}{2}+\frac{-2}{3}+\dots=\ln3\;$ is slightly less well known (I think)
$[\overline{1,0,-1,0}]=\frac{1}{1}+\frac{0}{2}+\frac{-1}{3}+\dots=\frac{\pi}{4}\;$ is the Gregory-Leibniz series for $\pi$

What I claim is that $[\overline{1,-1,-2,-1,1,2}]$ is equal to $0$. I wonder if there are any simple proofs of this (i.e. definitely without using calculus, preferably without appealing to complex numbers/taylor series/etc.)

P.S. I know a method that doesn't use any integrals or derivatives, but requires knowledge of the taylor series for $\ln(x)$ and the Euler formula for $e^{ix}$.

The reason I believe that there should be an elementary proof is that the sum, $0$, is a very simple number.

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  • $\begingroup$ I believe all of those "well known" harmonic-like sums have been proven using calculus. If so, then why do you expect there to be a non-calculus solution to evaluate this sum? $\endgroup$ – JimmyK4542 Jul 31 '14 at 2:14
  • $\begingroup$ Because the answer is 0. I feel that such a simple answer deserves a simple proof. $\endgroup$ – Akiva Weinberger Jul 31 '14 at 2:14
  • $\begingroup$ There is, I"m typing it up right now. $\endgroup$ – Semiclassical Jul 31 '14 at 2:16
  • $\begingroup$ ^Ahh, very well done. $\endgroup$ – JimmyK4542 Jul 31 '14 at 2:23
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    $\begingroup$ The overbar notation may be also found in davidhbailey.com/dhbpapers/digits.pdf and this series, as well as some others, in oeis.org/wiki/User:Jaume_Oliver_Lafont/Zero_relations. $\endgroup$ – Jaume Oliver Lafont Jan 8 '16 at 12:16
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We may rewrite your series in the following manner:

\begin{align} &\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)\\ &=\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{1}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{-1}{6n+6}\right)\\ &\hspace{1cm}-\sum_{n=0}^\infty\left(\frac{3}{6n+3}-\frac{3}{6n+6}\right)\\ \end{align} But these summations are both the alternating series $\sum_{n=0}^\infty \dfrac{(-1)^n}{n+1}$. Therefore they cancel and the summation is equal to zero.

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  • $\begingroup$ Ah. I wonder why I didn't notice that before! Good job. $\endgroup$ – Akiva Weinberger Jul 31 '14 at 2:24
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    $\begingroup$ Good answer. $\log(2)-\log(2)=0$ $\endgroup$ – robjohn Jul 31 '14 at 2:26
  • $\begingroup$ Looking at this again: We could also do a "$\log(3)-\log(3)$" proof, using the series for $\log(3)$ given in the question. In the notation introduced there:$$[\overline{1,-1,-2,-1,1,2}]=[\overline{1,1,-2,1,1,-2}]-[\overline{0,2,0,2,0,-4}],$$which equals $[\overline{1,1,-2}]-[\overline{1,1,-2}]=0$. $\endgroup$ – Akiva Weinberger May 29 '17 at 21:00
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In the language of Dirichlet series and the Riemann zeta function I believe this could be counted as an elementary proof:

Add the variable $s$ as an exponent to your series so that it becomes:

$$\sum_{n=0}^\infty\left(\frac{1}{(6n+1)^s}+\frac{-1}{(6n+2)^s}+\frac{-2}{(6n+3)^s}+\frac{-1}{(6n+4)^s}+\frac{1}{(6n+5)^s}+\frac{2}{(6n+6)^s}\right)$$

$$=\zeta(s)\left(1-\frac{1}{2^{s-1}}\right)\left(1-\frac{1}{3^{s-1}}\right) $$

In the case of $s=1$ we have exactly your series.

Therefore we investigate the limit:

$$\lim_{s\to 1} \, \zeta(s)\left(1-\frac{1}{2^{s-1}}\right)\left(1-\frac{1}{3^{s-1}}\right)$$

taking only parts of the limit we have:

$$\lim_{s\to 1} \, \zeta(s)\left(1-\frac{1}{2^{s-1}}\right)=\log(2)$$

and:

$$\lim_{s\to 1} \, \left(1-\frac{1}{3^{s-1}}\right)=0$$

therefore we have:

$$\lim_{s\to 1} \, \zeta(s)\left(1-\frac{1}{2^{s-1}}\right)\left(1-\frac{1}{3^{s-1}}\right)=\log(2) \cdot 0 = 0$$

hence:

$$\lim_{s\to 1} \, \sum_{n=0}^\infty\left(\frac{1}{(6n+1)^s}+\frac{-1}{(6n+2)^s}+\frac{-2}{(6n+3)^s}+\frac{-1}{(6n+4)^s}+\frac{1}{(6n+5)^s}+\frac{2}{(6n+6)^s}\right)=0$$

which is equivalent to: $$\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)=0$$

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    $\begingroup$ Wow. Very interesting. And I think I know how to transform this proof into @Semiclassical's; note that $\zeta(s)(1-2^{1-s})=\eta(s)$, the alternating zeta function, and then write that last limit out. This is interesting stuff. $\endgroup$ – Akiva Weinberger Apr 22 '15 at 16:37
  • $\begingroup$ Ok, just in case you would be interested in the alternating zeta function via matrices, I post you this link: math.stackexchange.com/a/547951/8530 $\endgroup$ – Mats Granvik Apr 22 '15 at 16:50
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I think we can "squeeze" something out of this: $$0=\sum_{n=0}^\infty\left(\frac{1}{6n+6}+\frac{-1}{6n+6}+\frac{-2}{6n+6}+\frac{-1}{6n+6}+\frac{1}{6n+6}+\frac{2}{6n+6}\right)\le\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)\le\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+1}+\frac{-2}{6n+1}+\frac{-1}{6n+1}+\frac{1}{6n+1}+\frac{2}{6n+1}\right)=0$$

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  • $\begingroup$ Ah. Never mind. The negative terms in the sum would reverse the inequalities for those terms. This will not work. $\endgroup$ – Laars Helenius Jul 31 '14 at 7:43
  • $\begingroup$ The previous answer gets my up vote then. $\endgroup$ – Laars Helenius Jul 31 '14 at 7:45
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This is a comment in response to Mats Granvik's answer.

I bet Mats Granvik's Trick generalizes to all Weinberger series!

Let $$f(s,\vec{a})=\sum_{n=1}^\infty {\frac{a_n}{n^s}} $$

be called a Weinberger series when $f(1,\vec{a})=0$

where $\vec{a}=(a_1, a_2,\dots a_p)$ and $\forall k\in \mathbb{N}, \text{ }a_{p+k}=a_{k}$.

Let $p=q_1^{e_1} \dots q_l^{e_l}$ be the prime factorization of $p$.

Conjecture $$f(1,\vec{a})=0\implies f(s,\vec{a})=\lim_{x\to s}\zeta(x)\prod_{i=1}^l(1-\frac{1}{q_i^{x-1}})^{e_i}$$

Let's consider another case. How about $\vec{a}=[1,-3,1,1]$?

$f(\vec{a},s)=\zeta(s)(1-\frac{1}{2^{s-1}})^2$

I think generally speaking the factor-ability of $f$ may rely on the series being a Weinberger series.

By the way this would mean that $p$ prime $\implies f(1, \vec{a}) \neq 0$ and I think this is the case based on proposition 13 of this paper.

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  • $\begingroup$ If this is quickly provable/{dismissable as false} I think we should answer it here and leave it but if it turns out to be a good question then I think I should rewrite this as a new question. $\endgroup$ – Mason Aug 15 '18 at 1:22
  • $\begingroup$ I think I may have to deal with constants too. So I can just demand that $a_1=1$ and that should account for this. $\endgroup$ – Mason Aug 15 '18 at 2:34
  • $\begingroup$ Needing to know what $s$ value zeros the function obviously has to do with how function factors... The sentence above looks silly to me 24 hours later. The next case to consider might be $[1,1,-5,1,1,-5,1,1,4]$. You can see a picture of these functions here. But I think I might be losing a lot of accuracy due to computational constraints. $\endgroup$ – Mason Aug 16 '18 at 19:19
  • $\begingroup$ I did follow up on this thought: math.stackexchange.com/questions/2897525/… $\endgroup$ – Mason Aug 28 '18 at 19:15

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