Re: Math Solutions

bugzpodder · **Joined:** Sun, 4 May 2003 16:04:19 UTC **Posts:** 2906

I am taking advantage of latex rendering engine on this board.

$Lemma:2\sqrt {n + 1} - 2\sqrt n < \frac{1}{{\sqrt n }} < 2\sqrt n - 2\sqrt {n - 1} \\ {\rm{Proof: 4}}n^2 + 4n < (2n + 1)^2 = 4n^2 + 4n + 1 \\ \sqrt {{\rm{4}}n^2 + 4n} < 2n + 1 \\$
$2\sqrt n \sqrt {n + 1} - 2\sqrt n \sqrt n < 1 \\ 2\sqrt {n + 1} - 2\sqrt n < \frac{1}{{\sqrt n }} \\ {\rm{Similarly, }}\frac{1}{{\sqrt n }} < 2\sqrt n - 2\sqrt {n - 1} \\$
$\sum\limits_{k = 1}^n {\left( {2\sqrt {k + 1} - 2\sqrt k } \right)} < \sum\limits_{k = 1}^n {\frac{1}{{\sqrt k }}} \\ 2\sqrt {n + 1} - 2 < \sum\limits_{k = 1}^n {\frac{1}{{\sqrt k }}} \\ {\rm{Also }}\sum\limits_{k = 1}^n {\frac{1}{{\sqrt k }}} = 1 + \sum\limits_{k = 2}^n {\frac{1}{{\sqrt k }}} < 1 + \sum\limits_{k = 2}^n {\left( {2\sqrt {k + 1} - 2\sqrt k } \right)} = 1 + 2\sqrt n - 2 = 2\sqrt n - 1 \\$

Soarer · **Posted:** Thu, 3 Jul 2003 16:31:49 UTC

There's a similar question in imo prelim hk.

Find
[1+1/sqrt2+1/sqrt3+...+1/sqrt80]
something like that

S.O.S. Mathematics CyberBoard

Re: Math Solutions

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