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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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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? –  JimmyK4542 Jul 31 '14 at 2:14
    
Because the answer is 0. I feel that such a simple answer deserves a simple proof. –  columbus8myhw Jul 31 '14 at 2:14
    
There is, I"m typing it up right now. –  Semiclassical Jul 31 '14 at 2:16
    
^Ahh, very well done. –  JimmyK4542 Jul 31 '14 at 2:23
    
Your series is equal to the Dirichlet series for MangoldtLambda[6]=0 joriki answered that here: math.stackexchange.com/a/51708/8530 and said that Abels theorem would be needed to make the proof rigourous. –  Mats Granvik Apr 22 at 13:33

3 Answers 3

up vote 18 down vote accepted

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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Ah. I wonder why I didn't notice that before! Good job. –  columbus8myhw Jul 31 '14 at 2:24
1  
Good answer. $\log(2)-\log(2)=0$ –  robjohn Jul 31 '14 at 2:26

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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Ah. Never mind. The negative terms in the sum would reverse the inequalities for those terms. This will not work. –  Laars Helenius Jul 31 '14 at 7:43
    
The previous answer gets my up vote then. –  Laars Helenius Jul 31 '14 at 7:45

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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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. –  columbus8myhw Apr 22 at 16:37
    
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 –  Mats Granvik Apr 22 at 16:50

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