Answer to Question #89348 in Algebra for Sakshi

Question

Prove ∑ni=11i√≤1n!√∏ni=2(i−1−−−−√)+2∑ni=21i√∑i=1n1i≤1n!∏i=2n(i−1)+2∑i=2n1i using Weierstrass inequality

Answer 1

We have to prove

$$\displaystyle\sum_{i=1}^n{1 \over \sqrt{i}}\leq {1 \over \sqrt{n!}}\displaystyle\prod_{i=2}^n(\sqrt{i-1})+2\displaystyle\sum_{i=2}^n{1 \over \sqrt{i}}$$

Consider

$$\displaystyle\sum_{i=1}^n{1 \over \sqrt{i}}=1+\displaystyle\sum_{i=2}^n{1 \over \sqrt{i}}$$

Subtract $(2\displaystyle\sum_{i=2}^n{1 \over \sqrt{i}})$ from both sides

$$\displaystyle\sum_{i=1}^n{1 \over \sqrt{i}}-2\displaystyle\sum_{i=2}^n{1 \over \sqrt{i}}=1+\displaystyle\sum_{i=2}^n{1 \over \sqrt{i}}-2\displaystyle\sum_{i=2}^n{1 \over \sqrt{i}}$$

$$\displaystyle\sum_{i=1}^n{1 \over \sqrt{i}}-2\displaystyle\sum_{i=2}^n{1 \over \sqrt{i}}=1-\displaystyle\sum_{i=2}^n{1 \over \sqrt{i}}$$

Hence we can rewrite the original inequality as

$$1-\displaystyle\sum_{i=2}^n{1 \over \sqrt{i}}\leq {1 \over \sqrt{n!}}\displaystyle\prod_{i=2}^n(\sqrt{i-1})$$

By the Weierstrass inequality (since ${1 \over \sqrt{i}}\isin (0, 1]$ for all $i \in\Z^+$) we have

$$1-\displaystyle\sum_{i=2}^n{1 \over \sqrt{i}}\leq \displaystyle\prod_{i=2}^n(1-{1 \over \sqrt{i}})= \displaystyle\prod_{i=2}^n({\sqrt i-1 \over \sqrt{i}})={1 \over \sqrt{n!}}\displaystyle\prod_{i=2}^n(\sqrt {i}-1)$$

For all $i \in\Z^+$

$$\sqrt{i}-\sqrt{i-1}={i-(i-1) \over {\sqrt{i}+\sqrt{i-1}}}\leq1$$

Then

$$\sqrt{i}-1\leq\sqrt{i-1}$$

Hence

$$1-\displaystyle\sum_{i=2}^n{1 \over \sqrt{i}}\leq{1 \over \sqrt{n!}}\displaystyle\prod_{i=2}^n(\sqrt {i}-1)\leq {1 \over \sqrt{n!}}\displaystyle\prod_{i=2}^n(\sqrt {i-1})$$

Therefore, we prove that

$$\displaystyle\sum_{i=1}^n{1 \over \sqrt{i}}\leq {1 \over \sqrt{n!}}\displaystyle\prod_{i=2}^n(\sqrt{i-1})+2\displaystyle\sum_{i=2}^n{1 \over \sqrt{i}}$$