# exponent of convergence of a sequence of complex numbers

#### pantboio

##### Member
Def. Let $\{z_j\}$ be a sequence of non-zero complex numbers. We call the exponent of convergence of the sequence the positive number $b$, if it exists,
$$b=inf\{\rho >0 :\sum_{j=1}^{+\infty}\frac{1}{|z_j|^{\rho}}<\infty \}$$

Now consider the function
$$f(z)=e^{e^z}-1$$
Find the zeros $\{z_j\}$ of $f$ and their exponent of convergence.

This is what i did: we have to solve the equation:
$$e^{e^z}=1$$
thus
$$e^z=2n_1\pi i$$
for $n_1$ integer not zero.
We have
$$z=log(2n_1\pi i)+2n_2\pi i$$
with $n_2$ integer.

Now i think the exponent is $\infty$, as we have a $log$, but i don't know how to formalize it. In particular, i have the problem to enumerate the $z_j's$, since each of them has two integers $n_1,n_2$ and i can't see how to express
$$\sum_{j=1}^{+\infty}\frac{1}{|z_j|}$$
Any help would be appreciated!

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#### pantboio

##### Member
I think i was completely wrong, indeed let $\{z_j\}$ the sequence of zeros of $f(z)=e^{e^z}-1$. We said that these points are of the type
$$z_{n_1,n_2}=log(2n_1\pi i)+2n_2\pi i$$
for $n_1,n_2$ integers, $n_1$ non-zero. Hence their modules are
$$\left((log|2n_1\pi|)^2+\frac{(4n_2+1)^2\pi^2}{4}\right)^{\frac{1}{2}}$$
which is asymptotic to $n_2$ for $n_1,n_2$ big enough. Therefore i think the convergence exponent to be 1$#### Opalg ##### MHB Oldtimer Staff member I think i was completely wrong, indeed let$\{z_j\}$the sequence of zeros of$f(z)=e^{e^z}-1$. We said that these points are of the type $$z_{n_1,n_2}=log(2n_1\pi i)+2n_2\pi i$$ for$n_1,n_2$integers,$n_1$non-zero. Hence their modules are $$\left((log|2n_1\pi|)^2+\frac{(4n_2+1)^2\pi^2}{4}\right)^{\frac{1}{2}}$$ which is asymptotic to$n_2$for$n_1,n_2$big enough. Therefore i think the convergence exponent to be 1$
I think you were right first time, to guess that the exponent is infinite. If you look at the subsequence of $\{z_{n_1,n_2}\}$ for which $n_2=0$, you see that $|z_{n_1,0}|\approx\log(2\pi n_1).$ That subsequence tends to infinity slower than any positive power of $n_1.$ In particular, $|z_{n_1,0}|< n_1^{1/\rho}$, from which it follows that $\displaystyle\sum_{n_1}\frac1{|z_{n_1,0}|^\rho} > \sum_{n_1}\frac1{n_1}$, which diverges. If a subsequence diverges then a fortiori the whole sequence diverges.