# Sum of a rearranged alternating harmonic series, with three positive terms followed by one negative term

The series is this:

$$1 + 1/3 + 1/5 - 1/2 + 1/7 + 1/9 + 1/11 - 1/4 + 1/13 + 1/15 + 1/17 - 1/6 ...$$

The hint is to consider partial sums to $4n$ terms. I did that, and got the summation

$$\sum 1/(6x-5) + 1/(6x-3) + 1/(6x-1) - 1/(2x)$$

But then I got stuck. The terms don't seem to simplify to anything... Wolfram alpha gives the result of this summation as $\log(12)/2$, but I have no idea how to get there.

Any help greatly appreciated.

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Come on, at least put them on a common denominator. –  Henning Makholm Nov 1 '11 at 23:04
I tried putting them on a common denominator, and got something crazy ie. $$(15-92 x+108 x^2)/(6 x (-1+2 x) (-5+6 x) (-1+6 x))$$... –  dhz Nov 1 '11 at 23:08

Hint:

Try writing the sum as $$\sum_{k=0}^\infty \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{3}{6k+6}\tag{1}$$ Then use the fact that \begin{align} \sum_{k=0}^n\frac{1}{2k+1} &=\log(2n+1)+\gamma-\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\\ &=\log(2)+\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\tag{2} \end{align} and that \begin{align} \sum_{k=0}^n\frac{1}{2k+2} &=\tfrac{1}{2}(\log(n+1)+\gamma)+O\left(\tfrac{1}{n}\right)\\ &=\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\tag{3} \end{align} Addendum:

Both formulas $(2)$ and $(3)$ follow from the asymptotic expansion for the partial harmonic series: $$\sum_{k=1}^n\frac{1}{k}=\log(n)+\gamma+O\left(\tfrac{1}{n}\right)$$ where $\gamma$ is the Euler-Mascheroni Constant. Thus, \begin{align} \sum_{k=0}^n\frac{1}{2k+1} &=\sum_{k=1}^{2n+1}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{2k}\\ &=\log(2n+1)+\gamma-\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\\ &=\log(2)+\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right) \end{align} and \begin{align} \sum_{k=0}^n\frac{1}{2k+2} &=\sum_{k=1}^{n+1}\frac{1}{2k}\\ &=\tfrac{1}{2}(\log(n+1)+\gamma)+O\left(\tfrac{1}{n}\right)\\ &=\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right) \end{align} Using $(2)$, we get \begin{align} \sum_{k=0}^n \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5} &=\log(2)+\tfrac{1}{2}(\log(3n+2)+\gamma)+O\left(\tfrac{1}{n}\right)\\ &=\tfrac{1}{2}(\log(n)+\log(12)+\gamma)+O\left(\tfrac{1}{n}\right)\tag{4} \end{align} and using $(3)$, we get \begin{align} \sum_{k=0}^n \frac{3}{6k+6} &=\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\tag{5} \end{align} Combining $(4)$ and $(5)$, we get $$\sum_{k=0}^n \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{3}{6k+6}=\tfrac{1}{2}\log(12)+O\left(\tfrac{1}{n}\right)$$

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A common trick used in combinatorics, but also useful here: Interpret $\displaystyle \frac{1}{k+1}$ as $\displaystyle \int^{1}_0 x^k dx.$

Then $$\sum_{k=1}^{\infty} \left( \frac{1}{6k-5} +\frac{1}{6k-3} + \frac{1}{6k-1} - \frac{1}{2k}\right) = \sum_{k=1}^{\infty} \int^1_0 \left( x^{6k-6} + x^{6k-2} + x^{6k} - x^{2k-1} \right) dx .$$

Now switch the sum and the integral, and simplify the inner terms by recognizing them as geometric series, and you will be left with the problem of integrating a rational function, which is routine.

EDIT: Henning's suggestion of putting them all on a common denominator reminded me of the method of summing certain series using the residue calculus, which works only if $f(n) = \mathcal{O}(1/n^2)$ which it is here. That tool, though more systematic, might be a bit heavy here considering the elementary solution.

EDIT 2: Sorry about that dhz, I should have included the details instead of just how to get the value.

First, let's clarify why it is sufficient to only consider the $4n$-th partial sums, $s_{4n}$. If $s_{4n} \to S$ then $s_{4n+1} = s_{4n} + \frac{1}{6n+1}$ so $s_{4n+1} \to S$ as well. $s_{4n+2}$ and $s_{4n+3}$ follow similarly, and it follows $s_n \to S.$

Now to avoid worrying about rearranging terms in a series which is not absolutely convergent, let us only consider $$s_{4n} =\sum_{k=1}^{4n} \left( \frac{1}{6k-5} +\frac{1}{6k-3} + \frac{1}{6k-1} - \frac{1}{2k}\right) = \sum_{k=1}^{4n} \int^1_0 \left( x^{6k-6} + x^{6k-2} + x^{6k} - x^{2k-1} \right) dx .$$

With this everything becomes clearly rigorous - interchanging a finite sum with an integral is always valid, so there's no need to justify that further, and we need not worry about rearranging any terms with harmful effects. The computation is only slightly more annoying, then take the limit to get the result.

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Hi Ragib, That's quite awesome, thanks! I didn't get the second part, though. After switching the sum and the integral, how would we be able to simplify the inner terms as a geometric series? In doing so, would we not have to rearrange the series again and hence get a different result than otherwise? If we don't rearrange, then I don't see how turning it into an integral helps? –  dhz Nov 1 '11 at 23:20
@dhz See my 2nd edit on why we don't have to worry =]. –  Ragib Zaman Nov 2 '11 at 4:28