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!
 A: 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!
