Answer to Question #89348 in Algebra for Sakshi

Question #89348
Prove ∑ni=11i√≤1n!√∏ni=2(i−1−−−−√)+2∑ni=21i√∑i=1n1i≤1n!∏i=2n(i−1)+2∑i=2n1i using Weierstrass inequality
1
Expert's answer
2019-05-08T08:19:53-0400

We have to prove


i=1n1i1n!i=2n(i1)+2i=2n1i\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


i=1n1i=1+i=2n1i\displaystyle\sum_{i=1}^n{1 \over \sqrt{i}}=1+\displaystyle\sum_{i=2}^n{1 \over \sqrt{i}}

Subtract (2i=2n1i)(2\displaystyle\sum_{i=2}^n{1 \over \sqrt{i}}) from both sides


i=1n1i2i=2n1i=1+i=2n1i2i=2n1i\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}}

i=1n1i2i=2n1i=1i=2n1i\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


1i=2n1i1n!i=2n(i1)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 1i(0,1]{1 \over \sqrt{i}}\isin (0, 1] for all iZ+i \in\Z^+) we have


1i=2n1ii=2n(11i)=i=2n(i1i)=1n!i=2n(i1)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 iZ+i \in\Z^+


ii1=i(i1)i+i11\sqrt{i}-\sqrt{i-1}={i-(i-1) \over {\sqrt{i}+\sqrt{i-1}}}\leq1

Then


i1i1\sqrt{i}-1\leq\sqrt{i-1}

Hence


1i=2n1i1n!i=2n(i1)1n!i=2n(i1)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


i=1n1i1n!i=2n(i1)+2i=2n1i\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}}

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Comments

Assignment Expert
08.05.19, 17:36

Dear Sakshi, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!

Sakshi
08.05.19, 17:28

Thanx alot for this solution and this fast service

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