Technique for constructing an entire function satisfying a given growth condition How can I construct an entire function whose growth rate at infinity satisfies 
$$ \lim_{r \to \infty} \frac{\log M(r)}{\sqrt {r}} =1$$
where $M(r) = \max_{|z|=r} |f(z)|$? 
Based on the above limit, I think that I can say $M(r)  \to   e^{\sqrt{r}}$, as r grows to infinity.  So, this shows that $f(z)$ has exponential growth like $e^\sqrt{r}$.  
Where can I go from here?  Can I manipulate the known series for $e^z$ and claim that this series 
$$\sum \frac{(\sqrt{z})^n}{n!}$$
is the entire function that we want?   
 A: Your idea goes in the right direction, but doesn't quite work out. If we choose a branch of $\sqrt{z}$ on a domain where one exists, and expand $e^{\sqrt{z}}$, we get
$$\sum_{n = 0}^\infty \frac{(\sqrt{z})^n}{n!} = \sum_{k = 0}^\infty \frac{(\sqrt{z})^{2k}}{(2k)!} + \sum_{k = 0}^\infty \frac{(\sqrt{z})^{2k+1}}{(2k+1)!} = \sum_{k = 0}^\infty \frac{z^k}{(2k)!} + \sqrt{z}\sum_{k = 0}^\infty \frac{z^k}{(2k+1)!}.$$
Both series clearly yield entire functions, but the $\sqrt{z}$ factor on the second sum makes it impossible for the whole thing to be an entire function. So what happens if we take only part of it? If we take the first series, we get
$$\sum_{k = 0}^\infty \frac{z^k}{(2k)!} = \cosh \sqrt{z} = \frac{1}{2}\bigl(e^{\sqrt{z}} + e^{-\sqrt{z}}\bigr).$$
That is an entire function (the ambiguity of $\sqrt{z}$ is annihilated by the evenness of $\cosh$/the fact that $\sqrt{z}$ and $-\sqrt{z}$ are both used in the same way), and the exponential representation strongly suggests - or makes it evident - that this function has the right growth behaviour. If we had taken the other series, the resulting function $\dfrac{\sinh \sqrt{z}}{\sqrt{z}}$ would also have the right growth, the $\sqrt{z}$ in the denominator is insignificant in comparison with the exponential involved in $\sinh$.
We can extend this approach systematically to obtain functions such that
$$\lim_{r\to \infty} \frac{\log M(r)}{r^\alpha} = 1$$
for rational $\alpha > 0$. Basically, we want something containing $e^{z^{\alpha}}$, but for $\alpha \notin \mathbb{N}$ we need a modification to obtain an entire function. If $\alpha = \frac{m}{k}$, we retain only the terms of
$$e^{z^\alpha} = \sum_{n = 0}^\infty \frac{(z^\alpha)^n}{n!}$$
for which the exponent $n \alpha$ is an integer, so $k \mid n$. That gives the function
$$g_{\alpha}(z) = \sum_{n = 0}^\infty \frac{(z^\alpha)^{kn}}{(kn)!} = \sum_{n = 0}^\infty \frac{z^{mn}}{(kn)!}.$$
To verify that this has the right growth behaviour, let $\zeta_k = \exp \frac{2\pi i}{k}$ and consider the function
$$h_k(z) = \frac{1}{k}\sum_{s = 0}^{k-1} e^{\zeta_k^s\cdot z} = \sum_{n = 0}^\infty \Biggl(\frac{1}{k}\sum_{s = 0}^{k-1} \zeta_k^{n\cdot s}\Biggr)\frac{z^n}{n!} = \sum_{n = 0}^\infty \frac{z^{kn}}{(kn)!}.$$
Then we see that $g_{\alpha}(z) = h_k({z^{\alpha}})$ has the right growth:
Since $\lvert e^{\zeta_k^s\cdot z}\rvert \leqslant e^{\lvert z\rvert}$, we have $\lvert h_k(z)\rvert \leqslant e^{\lvert z\rvert}$, and for $x > 0$ we have $\operatorname{Re} (\zeta_k^s x) = \bigl(\cos \frac{2\pi s}{k}\bigr)x \leqslant \bigl(\cos \frac{2\pi}{k}\bigr)x$ for  $0 < s < k$, so
$$\lvert h_k(x)\rvert \geqslant \frac{1}{k} e^{x} - \frac{k-1}{k} e^{x\cos \frac{2\pi}{k}} = \frac{e^x}{k}\Bigl(1 - (k-1)e^{-(1-\cos \frac{2\pi}{k})x}\Bigr) \sim \frac{e^x}{k}.$$
Altogether it follows that $\log M_{h_k}(r) \sim r$ and therefore $\log M_{g_\alpha}(r) \sim r^\alpha$.
