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I would like to ask how to prove the inequality: $\sum_{n=0}^{\infty}\frac{|z|^n}{(2n)!}\leq\sum_{n=0}^{\infty}\frac{(|z|^\frac{1}{2})^n}{n!}$ where $z\in\mathbb{C}$.

I find this inequality in the steps of proving $\left|\cos\left(z^\frac{1}{2}\right)\right|\leq e^{|z|^\frac{1}{2}}$.

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    $\begingroup$ You are asked to prove $\cosh(\sqrt{|z|})\leq e^{\sqrt{|z|}}$, which is trivial, equivalent to $\sinh\sqrt{|z|}\geq 0$. $\endgroup$ Jun 23 at 14:49

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Writing the right-hand-side as the sum of the even part and the odd part: $$\sum_{n=0}^{\infty}\frac{(|z|^\frac{1}{2})^n}{n!}=\sum_{k=0}^{\infty}\frac{(|z|^\frac{1}{2})^{2k}}{(2k)!}+\sum_{k=0}^{\infty}\frac{(|z|^\frac{1}{2})^{2k+1}}{(2k+1)!}\geq \sum_{k=0}^{\infty}\frac{(|z|^\frac{1}{2})^{2k}}{(2k)!}=\sum_{k=0}^{\infty}\frac{|z|^k}{(2k)!},$$ i.e., $$\sum_{n=0}^{\infty}\frac{(|z|^\frac{1}{2})^n}{n!}\geq \sum_{k=0}^{\infty}\frac{|z|^k}{(2k)!}=\sum_{n=0}^{\infty}\frac{|z|^n}{(2n)!}.$$

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  • $\begingroup$ Thank you, I get it now. $\endgroup$
    – math noob
    Jun 23 at 14:22

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