Radius of convergence? Can someone tel me how to calculate the radius of convergence of this, power serie :
$\sum_{n>0} \frac{z^{n!}}{n}$
Thanks for any answer
 A: In a general way, the radius of convergence of the series
$$
\sum_{k > 0} a_k z^k
$$
is the limit sup of $\lvert a_k\rvert ^{1/k}$. In your problem one has $a_k = l$ for $k = l!$ (with $l > 0$) and $a_k = 0$ otherwise.
Clearly the sequence $(a_k)_k$ cannot be bounded by a geometric sequence with ratio smaller than $1$, because if it were, it would tend to $0$, which is not the case (obviously $(a_k)_k$ is unbounded). Therefore $\limsup_{k \to \infty} \lvert a_k\rvert ^{1/k} \geq 1$ (this follows from reasoning by contraposition, observing that any sequence of nonnegative real numbers $(u_k)_k$ such that $\limsup u_k^{1/k} \leq \lambda$ can be bounded by a multiple of $(\lambda')^k$ for any $\lambda' > \lambda$.
Conversely, since $l! \geq 2^{l - 1}$ for all $l$, $k = l!$ implies that $l \leq \log_2 k  + 1$, so $\lvert a_k \rvert \leq \log_2 k + 1$ for all $k$. But the sequence $(\log_2 k + 1)_k$ grows slowlier than any exponential (actually even slowlier than any polynomial), so $\limsup \lvert a_k^{1/k}\rvert \leq 1$.
We conclude that the wanted radius of convergence is $1$.
Regards, /Nancy-N
