Order of the entire function: $\prod\limits_{n=1}^{\infty} \left(1-\frac{z}{n^k}\right)$ Please how to find order of $$ f_k(z) = \prod\limits_{n=1}^{\infty} \left(1-\frac{z}{n^k}\right) .$$  
Let $M(r) = \max \{|f_k(z)|:|z| = r\}.$ Then order of $f_k(z)$ is defined as : $$\lambda = \limsup_{r\to \infty} \frac{\log \log M(r)}{\log r}.$$  
Can someone help to solve this. Thank you.
 A: The order of this function is $\frac{1}{k}$, and we must have $k>1$.
Proof: First, note that $\prod_{n=1}^{\infty}\left(1-\frac{z}{n^{k}}\right)$ will attain its maximum modulus on the circle of radius $r$ at $z=-r$.  Taking the logarithm of this, we need to examine
$$h(r)=\log\prod_{n=1}^{\infty}\left(1+\frac{r}{n^{k}}\right)=\sum_{n=1}^{\infty}\log\left(1+\frac{r}{n^{k}}\right).$$  If we can prove that this is bounded below and above by something close to $r^\frac{1}{k}$, then we will have shown that $\lambda=\frac{1}{k}$ since $$\lambda = \lim_{r\rightarrow \infty} \frac{\log h(r)}{\log r}.$$ 
Lower Bound: We  have that $$\sum_{n=1}^{\infty}\log\left(1+\frac{r}{n^{k}}\right)\geq\sum_{n\leq r^{\frac{\epsilon}{k}}}\log\left(1+\frac{r}{n^{k}}\right)\geq r^{\frac{\epsilon}{k}}\log\left(1+r^{1-\epsilon}\right)$$ for any $0<\epsilon<1.$ Taking $\epsilon=1-\frac{1}{\log r},$ we see that $r^{1-\epsilon}=e,$ and hence the above is $$\geq e^{-\frac{1}{k}}r^{\frac{1}{k}}\log\left(1+e\right)\geq e^{-\frac{1}{k}}r^{\frac{1}{k}}.$$ This then implies that $\lambda\geq\frac{1}{k}.$ 
Upper Bound:  We split the sum based on the size of $n$.  If $n>r^{\frac{1}{k}},$ by using the bound $\log\left(1+\frac{r}{n^{k}}\right)\leq\frac{r}{n^{k}},$ we have $$\sum_{n>r^{\frac{1}{k}}}^{\infty}\log\left(1+\frac{r}{n^{k}}\right)\leq r\sum_{n>r^{\frac{1}{k}}}^{\infty}\frac{1}{n^{k}}\ll\frac{r}{r^{\frac{1}{k}k-1}}=r^{\frac{1}{k}}.$$  Next, for $n^{k}\leq r,$ we have that $\log\left(1+\frac{r}{n^{k}}\right)\leq\log\left(r+1\right),$ which implies that $$\sum_{n\leq r^{\frac{1}{k}}}^{\infty}\log\left(1+\frac{r}{n^{k}}\right)\leq r^{\frac{1}{k}}\log\left(r+1\right).$$ Thus $$\sum_{n=1}^{\infty}\log\left(1+\frac{r}{n^{k}}\right)\ll r^{\frac{1}{k}}\log r,$$ which implies that $\lambda\leq\frac{1}{k}.$
