Given the powerseries $\sum_{k=1}^{\infty} a_k x^k$, where $a_k = \sum_{j=1}^{k} \frac{1}{j!} \binom{k-1}{j-1}$, I want to find the convergence radius.
This however I find quite difficult considering the $a_k$ coefficient is given as a sum and not just a single term. I know the $\lim\limits_{k \to \infty} \left|\frac{a_k}{a_{k+1}}\right|$ and $\left(\limsup\limits_{k \to \infty} |a_k|^{1/k}\right)^{-1}$ criterions, but can't seem to get them to work with sums..
Any help would really be appreciated
Since$$(\forall j\in\mathbb N):\left(\frac z{1-z}\right)^j=\sum_{k=j}^{+\infty}\binom{k-1}{j-1}z^k$$we have:\begin{align}\exp\left(\frac z{1-z}\right)&=\sum_{j=0}^\infty\frac1{j!}\left(\frac z{1-z}\right)^j\\&=1+\sum_{j=1}^{+\infty}\frac1{j!}\sum_{k=j}^{+\infty}\binom{k-1}{j-1}z^k\\&=1+\sum_{k=1}^{+\infty}\left(\sum_{j=1}^k\frac1{j!}\binom{k-1}{j-1}\right)z^k.\end{align}So, the radius of convergence is at least $1$, since this Taylor series must converge on any open disk centered on $0$ which is contained in $\mathbb C\setminus\{1\}$. But it cannot be greater than $1$, because otherwise the limit$$\lim_{z\to1}1+\sum_{k=1}^{+\infty}\left(\sum_{j=1}^k\frac1{j!}\binom{k-1}{j-1}\right)z^k$$would exist. But it doesn't exist, since the limit$$\lim_{z\to1}\exp\left(\frac z{1-z}\right)$$doesn't exist. So, the radius of convergence is $1$.