# Prove $\sum_{i=1}^n\frac{1}{\sqrt{i}}\le\frac{1}{\sqrt{n!}}\prod_{i=2}^n(\sqrt{i-1})+2\sum_{i=2}^n\frac{1}{\sqrt{i}}$ using Weierstrass inequality

Use Weierstrass's inequality to prove that $$\sum_{i=1}^n \frac{1}{\sqrt{i}}\le \frac{1}{\sqrt{n!}} \prod_{i = 2}^n (\sqrt{i-1}) + 2 \left(\sum_{i = 2}^n \frac{1}{\sqrt{i}}\right).$$

Using Weierstrass's inequality, I get

$$1 + \sum_{i=1}^n \frac{1}{\sqrt{i}} \le \frac{1}{\sqrt{n!}} \left(\prod_{i = 1}^n (1 + \sqrt{i})\right)$$

• The very same question has already been asked here, so please use approach0 to search before you ask, to see if your question (likely a homework problem if two people ask it in such a short time frame) has been ask before. – Viktor Glombik Mar 21 at 18:09

Hint: Since $$\sum_{k = 1}^{n} \frac{1}{\sqrt{k}} = 1 + \sum_{k = 2}^{n} \frac{1}{\sqrt{k}},$$ we can rewrite the inequality, which has to be shown (by subtracting $$2 \sum_{k = 2}^{n} \frac{1}{\sqrt{k}}$$ from both sides) as $$1 - \sum_{k = 2}^{n} \frac{1}{\sqrt{k}} \le \frac{1}{\sqrt{n!}} \prod_{k = 2}^n (\sqrt{k-1}).$$ By the Weierstrass inequality (since $$\frac{1}{\sqrt{k}} \in (0,1]$$ for all $$k \in \mathbb{N}_{>0}$$) we have $$1 - \sum_{k = 2}^{n} \frac{1}{\sqrt{k}} \le \prod_{k = 2}^{n} \left(1 - \frac{1}{\sqrt{k}}\right) = \prod_{k = 2}^{n} \frac{\sqrt{k} - 1}{\sqrt{k}} = \frac{1}{\sqrt{n!}} \prod_{k = 2}^{n} (\sqrt{k}-1), \le \frac{1}{\sqrt{n!}} \prod_{k = 2}^{n} \sqrt{k - 1}$$ where the last equality is achieved by pulling out the denominator out of the product and the last inequality is using the fact that $$\sqrt{k}-1 \le \sqrt{k - 1}$$ for all $$k \in \mathbb{N}_{>1}$$, which you can easily verify either by induction or by drawing a graph.