# When the Plethystic sum converges?

We have \begin{align} & \exp\left(-\sum_{p=1}^{\infty} \frac{x^p}{p}\right) = 1-x, \\ & \exp\left(-\sum_{p=1}^{\infty} \frac{\sum_{k} x_k^p}{p}\right) = \prod_{k}(1 - x_k). \end{align} When do $\exp\left(-\sum_{p=1}^{\infty}\frac{x^p}{p}\right)$ and $\exp\left(-\sum_{p=1}^{\infty} \frac{\sum_{k} x_k^p}{p}\right)$ converge?

Any help will be greatly appreciated!

We can find the interval of convergence of the power series $S(x)=\sum_{p=1}^\infty \frac{x^p}{p}$ using a standard tool, namely the ratio test. Applying the ratio test, we find that the series converges for $|x|<1$ and diverges for $|x|>1$.

For $x=1$ the series is the Harmonic series, which diverges. For $x=-1$, the series is the alternating harmonic series, which converges to $-\log(2)$. Therefore, the interval of convergence is $-1\le x<1$.

Next, we will show that for $-1\le x<1$, the series

$$\sum_{p=1}^\infty \frac{x^p}{p}=-\log(1-x)$$

To show this, we write for $|x|<1$

\begin{align} \sum_{p=1}^N \frac{x^p}{p}&=\sum_{p=1}^N \int_0^x t^{p-1}\,dt\\\\ &=\int_0^x \sum_{p=1}^Nt^{p-1}\\\\ &=\int_0^x\frac{1-t^N}{1-t} \,dt\\\\ &=-\log(1-x)-\int_0^x\frac{t^N}{1-t}\,dt \end{align}

It is easy to show (e.g., Apply the Dominated Convergence Theorem, or simply integrate by parts with $u=1/(1-t)$ and $v=t^{N+1}/(N+1)$) that for $-1\le x<1$,

$$\lim_{N\to \infty}\int_0^x\frac{t^N}{1-t}\,dt=0$$

whereby we conclude immediately for $-1\le x<1$

$$\sum_{p=1}^\infty \frac{x^p}{p}=-\log(1-x)$$

as was to be shown!