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Playing around with Maple I found this identity $$\sum_{k=0}^{n-1}\frac{2k+1}{1-z^{2k+1}}=n\sum_{k=0}^{n-1}\frac{1}{1+z^{k}}$$ where $n$ is a positive integer, $z=\exp(\pi i/n)$. I was able to verify it only numerically. Does anyone know how to prove it?

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  • $\begingroup$ Are you sure? All the expressions $1/(1+z^k)$ and $1/(1-z^{2k+1})$ have real parts equal to $1/2$ but the coefficients on the left add up to more than $n.$ $\endgroup$ Dec 18, 2015 at 19:40
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    $\begingroup$ @Justpassingby: The sum of the coefficients on the right is $n^2$, not $n$ (note the $n$ in front of the sum). That matches the sum of the coefficients on the left. So now we have proved that the real parts match! $\endgroup$ Dec 18, 2015 at 19:46
  • $\begingroup$ @Henning Quite so. That establishes the real part of the equality then :-) $\endgroup$ Dec 18, 2015 at 19:49
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    $\begingroup$ The imaginary parts then boil down to $$ \sum_{k=0}^{n-1} (2k+1)\tan\left((2k+1-n)\tfrac\pi{2n}\right) = n \sum_{k=0}^{n-1} \tan\left(k\tfrac\pi{2n}\right)$$ $\endgroup$ Dec 18, 2015 at 20:02
  • $\begingroup$ The RHS has imaginary part 0 baecause the terms for k and n-k cancel out (for k=0 it is real). $\endgroup$ Dec 19, 2015 at 10:27

3 Answers 3

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First we observe that:

$$\sum_{k=1}^{2n-1} {k\over {1-z^k}} - \sum_{k=1}^{n-1} {2k\over {1-z^{2k}}} = \sum_{k=1}^{n-1} {2k+1 \over {1-z^{2k+1}}} $$

Now, by partial fractions,

$${2k \over {1-z^{2k}}} = {k\over {1-z^k}} + {k\over {1+z^k}} $$ Hence:

$$\sum_{k=1}^{2n-1} {k\over {1-z^k}} - \bigg( \sum_{k=1}^{n-1} {k\over {1-z^{k}}} + \sum_{k=1}^{n-1} {k\over {1+z^{k}}}\bigg) = \sum_{k=1}^{n-1} {2k+1 \over {1-z^{2k+1}}} $$ Now, by combining the first two sums:

$$\sum_{k=n}^{2n-1} {k\over {1-z^k}} - \sum_{k=1}^{n-1} {k\over {1+z^{k}}} = \sum_{k=1}^{n-1} {2k+1 \over {1-z^{2k+1}}} $$

Now let's re-write the first sum:

$$\sum_{k=0}^{n-1} {k+n\over {1-z^{k+n}}} - \sum_{k=0}^{n-1} {k\over {1+z^{k}}} = \sum_{k=1}^{n-1} {2k+1 \over {1-z^{2k+1}}} $$

Now let's use the fact that $z^n=-1$, and combine the final two terms:

$$\sum_{k=0}^{n-1} {n\over {1+z^{k}}} = \sum_{k=1}^{n-1} {2k+1 \over {1-z^{2k+1}}} $$

As required.

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    $\begingroup$ Smart answer. Very nice! +1 $\endgroup$ Dec 23, 2015 at 21:41
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Given positive integer $n$ that defines $z := \exp(\pi i/n),$ we can generalize the oddly periodic sum with $$ S_n(x) := \sum_{k=0}^{n-1}\frac{2k+1}{1-xz^{2k+1}}. $$ The power series when $|x| < 1$ is $$ S_n(x) = \sum_{E=0}^{\infty}x^E\sum_{k=0}^{n-1}(2k+1)z^{2kE+E}. $$ This simplifies with the weighted sum, $$ \left.\frac{d}{dr}(\frac{r-r^{2n+1}}{1-r^2})\right|_{r=z^E} = \sum_{k=0}^{n-1}(2k+1)z^{2kE},\ \mbox{ to} $$ $$ S_n(x) = \sum_{E=0}^{\infty}\frac{1-z^{2E}-(2n+1)z^{2En}+(2n+1)z^{2En+2E}+ 2z^{2E}-2z^{2En+2E}}{(1-z^{2E})^2}(xz)^E. $$ We observe that $z^{2E}\to1$ correctly reduces the fraction to $n^2,\ $ the period is (almost) $n,\ $ and $$ S_n(x) = \frac{1}{1+x^n}\left( n^2 - 2n\sum_{E=1}^{n-1} \frac{(xz)^E}{1-z^{2E}} \right). $$ Before we take $x\to1,\ $ we need to recall $$ z=\exp(\pi i/n)\ \implies\ n = 1+\sum_{E=1}^{n-1}\frac{2}{1-z^{2E}}. $$ We can evaluate our analytic sum at $$ S_n(1) = \frac{n}{2}\left(1 + 2\sum_{E=1}^{n-1}\frac{1-z^E}{1-z^{2E}} \right) $$ to find the desired relationship $$ \sum_{k=0}^{n-1}\frac{2k+1}{1-z^{2k+1}}\ =\ \sum_{k=0}^{n-1}\frac{n}{1+z^k}. $$

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Your identity is true. I give here the method for you to prove the point for general $n$.

One has as main preliminary remark $$\color{red}{z^n=-1\iff z^{n+k}=-z^k\iff \frac{1}{1-z^k}=\frac{z^{n-k}}{1+z^{n-k}}}$$ We make $$A=\sum_{k=o}^{n-1}\frac{2k+1}{1-z^{2k+1}}$$ $$B=n\sum_{k=0}^{n-1}\frac{1}{1+z^k}$$

Example of algebraic verification: $n=6$, even.

The odd exponents in $A$ are $\begin{cases}1\\3\\5\\7=6+1\\9=6+3\\11=6+5\end{cases}$ therefore $$A=\left({1\over 1-z}+{3\over 1-z^3}+{5\over 1-z^5}\right)+\left({6+1\over 1+z}+{6+3\over 1+z^3}+{6+5\over 1+z^5}\right)$$ $$A=B\Rightarrow \left({1\over 1-z}+{3\over 1-z^3}+{5\over 1-z^5}\right)+\left({1\over 1+z}+{3\over 1+z^3}+{5\over 1+z^5}\right)= 6\left({1\over 1+1}+{1\over 1+z^2}+{1\over 1+z^4}\right)$$ Hence $${2\over 1-z^2}+{6\over 1+1}+{10\over 1-z^{10}}= {6\over 1+1}+{6\over 1+z^2}+{6\over 1+z^4}$$ $${2\over 1-z^2}+{10\over 1+z^4}= {6\over 1+z^2}+{6\over 1+z^4}\iff {1\over 1+z^4}={1-2z^2\over 1-z^4}\iff z^4-z^2+1=0 $$ Since $z^6+1=(z^2+1)(z^4-z^2+1)=0$ the algebraic verification is ended.

Example of algebraic verification: $n=7$, odd.

$$A={1\over 1-z}+{3\over 1-z^3}+{5\over 1-z^5}+{7\over 1+1}+{7+2\over 1+z^2}+{7+4\over 1+z^4}+{7+6\over 1+z^6}=B$$ $$\left({1-6z\over 1-z}-{7\over 1+z}+{7z^2\over 1-z^2}+{2+5z^2\over 1+z^2}+{4\over 1+z^4}\right)={7\over 1+z^3}-{3\over 1-z^3}$$ $$\left({2z(2z^7-z^6+z^5-z^4+6z^3-z^2+z-1)\over z^8-1}\right)={7\over 1+z^3}-{3\over 1-z^3}$$ but the parenthesis equals $${2z(5z^3-2)\over –z-1}$$ because $$\left({2z(2z^7-(\color{red}{z^6-z^5+z^4-z^3+z^2-z+1})+5z^3)\over z^8-1}\right)={2z(5z^3-2)\over –z-1}$$ where the red polynomial is null as a factor of $z^7+1=0$. Therefore $${2z(5z^3-2)\over –z-1}={7\over 1+z^3}-{3\over 1-z^3}= {2(5z^2-2)\over z^6-1}$$ i.e $${z\over –z-1}={1\over z^6-1}\iff z^7-z=-z-1$$ which ends the proof $n=7$.

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