# Check whether $\sum\limits_{n=1}^{\infty}\frac{z^n}{(1-z^n)^k}=\sum\limits_{n=1}^{\infty}\sigma_{k-1}(n)z^n$

Is it true that

$$\sum_{n=1}^{\infty}\frac{z^n}{(1-z^n)^k}=\sum_{n=1}^{\infty}\sigma_{k-1}(n)z^n$$

If yes, how can I prove it?

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What is $\large\sigma_{\,k - 1}\left(n\right)$ ? –  Felix Marin Sep 29 '13 at 5:47
$\sigma_{k-1}(n)=\sum_{d|n}d^{k-1}$ –  user95733 Sep 29 '13 at 5:49
$\displaystyle{\large\sigma_{k - 1}\left(n\right) \equiv \sum_{d|n}d^{k-1}}$. Thanks. –  Felix Marin Sep 29 '13 at 5:58
It is not true. –  Ethan Sep 30 '13 at 5:44
I thought you were asking this for every $k$. Why are you suddenly interested in $k=1$ and $k=2$ only? –  Did Oct 3 '13 at 14:30

One can start from the identity $$\sum_{n=1}^{\infty}\frac{z^n}{(1-z^n)^k}=\sum_{n\geqslant1}z^n\sum_{i_1\geqslant0}z^{ni_1}\cdots\sum_{i_k\geqslant0}z^{ni_k}=\sum_{n,i_1,\ldots,i_k}z^{n(1+i_1+\cdots+i_k)}=\sum_{N\geqslant1}c_Nz^N,$$ where $c_N$ enumerates the $(k+1)$uples $(n,i_1,\ldots,i_k)$ such that $n$ is positive, every $i_j$ is nonnegative and $n(1+i_1+\cdots+i_k)=N$.

Thus, $n$ divides $N$ hence $d=N/n$ divides $N$. For every $d$, the number of $k$uples $(i_1,\ldots,i_k)$ such that $i_1+\cdots+i_k=d-1$ and every $i_j$ is nonnegative is the coefficient of $x^{d-1}$ in the series $$\left(\sum_{i\geqslant0}x^i\right)^k=\frac1{(1-x)^k}=\sum_{m\geqslant0}{m+k-1\choose k-1}x^m.$$ This shows that, for every $N\geqslant1$, $$c_N=\sum_{d\mid N}{d+k-2\choose k-1}.$$ One sees that $(c_N)_N$ is not the sequence $(\sigma_{k-1}(N))_N$ except if $k=1$ or $k=2$, for example, if $k=3$, $$c_N=\sum_{d\mid N}\tfrac12d(d+1)=\tfrac12\sigma_2(N)+\tfrac12\sigma_1(N).$$ More generally, $c_N$ is a barycenter of the coefficients $(\sigma_{i}(N))_{1\leqslant i\leqslant k-1}$. For every $k\geqslant1$, $c_1=1$, $c_2=1+k$, $c_3=1+\frac12k(k+1)$, $c_4=1+k+\frac16k(k+1)(k+2)$.

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$\displaystyle{% \sum_{n = 1}^{\infty}{z^{n} \over \left(1 - z^{n}\right)^{k}} = \sum_{n = 1}^{\infty}\sigma_{\,k - 1}\left(n\right)z^{n}:\ {\large ?} \quad\mbox{where}\quad \sigma_{\,k - 1}\left(n\right) \equiv \sum_{d\,|\,n}d^{k - 1}}$

With $\left\vert z\right\vert < 1$:

$${1 \over 1 - z} = \sum_{\ell = 0}^{\infty}z^{\ell}\,, \quad {1 \over \left(1 - z\right)^{2}} = \sum_{\ell = 1}^{\infty}\ell\,z^{\ell - 1}\,, \quad {1 \over \left(1 - z\right)^{3}} = {1 \over 2} \sum_{\ell = 2}^{\infty}\ell\left(\ell - 1\right)z^{\ell - 2}$$

$${1 \over \left(1 - z\right)^{4}} = {1 \over 3\cdot 2} \sum_{\ell = 3}^{\infty}\ell\left(\ell - 1\right)\left(\ell - 2\right) z^{\ell - 3}\,, \quad \ldots$$

$$\begin{array}{rcl}\hline\\ {1 \over \left(1 - z\right)^{k}} & = & {1 \over \left(k - 1\right)!} \sum_{\ell = k - 1}^{\infty}\ell\left(\ell - 1\right)\ldots \left(\ell - k + 2\right)z^{\ell - k + 1} \\[3mm]& = & {1 \over \left(k - 1\right)!} \sum_{\ell = k - 1}^{\infty}{\ell! \over \left(\ell - k + 1\right)!}\, z^{\ell - k + 1} \\[3mm]& = & \sum_{\ell = k - 1}^{\infty}{\ell \choose k - 1}z^{\ell - k + 1} = \sum_{\ell = 0}^{\infty}{\ell + k - 1\choose k - 1}z^{\ell} \\ \\ \hline \end{array}$$

Then,

\begin{align} \color{#ff0000}{\large\sum_{n = 1}^{\infty}{z^{n} \over \left(1 - z^{n}\right)^{k}}} &= \sum_{n = 1}^{\infty}z^{n}\sum_{\ell = 0}^{\infty}{\ell + k - 1\choose k - 1} z^{n\ell} = \sum_{\ell = 0}^{\infty}{\ell + k - 1 \choose k - 1} \sum_{n = 1}^{\infty}z^{n\ell + n} \\[3mm]&= \sum_{\ell = 0}^{\infty}{\ell + k - 1\choose k - 1} {z^{\ell + 1} \over 1 - z^{\ell + 1}} \\[3mm]&= \sum_{\ell = 0}^{\infty}{\ell + k - 1 \choose k - 1} z^{\ell + 1}\sum_{\ell' = 0}^{\infty}\left(z^{\ell + 1}\right)^{\ell'} \sum_{n = 1}^{\infty}\delta_{n, \ell + \ell' + \ell\ell' + 1} \\[3mm]&= \sum_{n = 1}^{\infty}z^{n} \sum_{\ell = 0}^{\infty}{\ell + k - 1 \choose k - 1} \sum_{\ell' = 0}^{\infty}\delta_{n,\ell + \ell' + \ell\ell' + 1} \\[3mm]&= \sum_{n = 1}^{\infty}z^{n} \sum_{\ell = 1}^{\infty}{\ell + k - 2 \choose k - 1} \sum_{\ell' = 1}^{\infty}\delta_{n,\ell\ell'} = \color{#ff0000}{\large\sum_{n = 1}^{\infty}z^{n} \sum_{\ell = 1 \atop \ell\,|\,n}^{n}{\ell + k - 2 \choose k - 1}} \end{align}

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This identity is true only for $k=1,2$ and for $|z|<1$. You can use this theorem.

Sum by Curves Theorem. An absolutely convergent series $\sum a_{mn}$ itself converges. Moreover, If $S_1\subseteq S_2\subseteq S_3\subseteq ...\subseteq\mathbb{N}\times\mathbb{N}$ is a nondecreasing sequence of finite sets having the property that for any $m,n$ there is a $\ell$ such that $$\mathbb{N}_m\times\mathbb{N}_n\subseteq S_\ell\subseteq S_{\ell+1}\subseteq S_{\ell+2}\subseteq ...,$$ then the sequence $s_\ell$ convergence, where $s_\ell$ is the finite sum $s_\ell:=\sum_{(m,n)\in S_\ell}a_{mn},$ and furthermore $\sum a_{mn}=\lim s_\ell.$

Let $S_\ell=T_1\bigcup...\bigcup T_\ell, T_\ell=\lbrace(m,n)\in\mathbb{N}\times\mathbb{N};mn=\ell\rbrace$

for $k=1$ note that $$\frac{1}{1-z^n}=\sum_{m=0}^{\infty}z^{mn}=\sum_{m=1}^{\infty}z^{n(m-1)}$$ $$\sum_{n=1}^{\infty}\frac{z^n}{1-z^n}=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}z^{mn}=\sum_{n=1}^{\infty}\sigma_0(n)z^n.$$

for $k=2$ note that $$\frac{1}{(1-z^n)^2}=\sum_{m=1}^{\infty}mz^{n(m-1)}$$ $$\sum_{n=1}^{\infty}\frac{z^n}{(1-z^n)^2}=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}mz^{mn}=\sum_{n=1}^{\infty}\sigma_1(n)z^n.$$

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What you can prove is that $$\sum_{n=1}^\infty \sigma_k(n)z^n = \sum_{m=1}^\infty f_k(z^m)$$ where $f_k(u) = \left(u\dfrac{d}{du}\right)^k \dfrac{u}{1-u}$. For $k\in\{0,1\}$ this is your formula, but not anymore for $k \geq 2$.

Proof. $$\sum_{n=1}^\infty \sigma_k(n)z^n = \sum_{n=1}^\infty \sum_{r \mid n}r^k z^n = \sum_{r=1}^\infty \sum_{m=1}^\infty r^k z^{rm} = \left(z\frac{d}{dz}\right)^k\sum_{r=1}^\infty \sum_{m=1}^\infty \frac{z^{rm}}{m^k}$$ and $$\sum_{r=1}^\infty \sum_{m=1}^\infty \frac{z^{rm}}{m^k} = \sum_{m=1}^\infty \frac{1}{m^k}\frac{z^m}{1-z^m}.$$

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