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 Post subject: Re: Math Solutions
PostPosted: Thu, 3 Jul 2003 16:18:32 UTC 
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I am taking advantage of latex rendering engine on this board. :D

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 \\

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Has anyone noticed that the below is WRONG? Otherwise this statement would be true:
-1\cong1\pmod{13}
i\cong5 \pmod{13} where
i^2=-1


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PostPosted: Thu, 3 Jul 2003 16:31:49 UTC 
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Joined: Fri, 2 May 2003 16:33:24 UTC
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Location: Hong Kong
There's a similar question in imo prelim hk.

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

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