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        riddles >> putnam exam (pure math) >> ***Spoiler*** inequality
        
(Message started by: anonymous on Jun 23^rd, 2003, 4:11pm)

Title: ***Spoiler*** inequality
Post by anonymous on Jun 23^rd, 2003, 4:11pm

\[
\begin{split}
\sum_{n=1}^{\infty} (a_n-a_{n+1})a_{n+2} \\
= \sum_{n=1}^{\infty} (a_1^{2^{n-1]-a_1^{2^n})a_1^{2^{n+1] \\
\leq \sum_{i=1}^{\infty} (a_1^i-a_1^{i+1})a_1^{2i+2} \\
= (1-a_1)a_1^2\sum_{i=1}^{\infty} a_1^{3i} \\
= (1-a_1)a_1^5/(1-a_1^3) \\
= a_1^5/(1+a_1+a_1^2) \\
< 1/3
\end{split}
\]

The second to third line uses telescopic sum, and there is a geometric series in the fourth line.

Title: Re: ***Spoiler*** inequality
Post by wowbagger on Jun 24^th, 2003, 5:17am

For those who aren't used to reading LaTeX:
http://www.ai.rug.nl/~towr/PHP/FORMULA/formula.php?md5=d33e933b365851c649bc1b0204d9c49c

Title: Re: ***Spoiler*** inequality
Post by James Fingas on Jun 24^th, 2003, 9:26am

How do we justify the third line? It's not true that every term in the new sum is larger than the corresponding term in the old sum. For instance, with i=n=2 and a₁=0.99, the terms are 0.017997 for the old and 0.009415 for the new.

I am assuming that you have a good justification ... maybe I don't understand telescopic sums like I thought I did?

Title: Re: ***Spoiler*** inequality
Post by anonymous on Jun 24^th, 2003, 10:27am

I'll take n=3 as an example

when n=3,
\[
\begin{split}
(a^{2^{n-1]-a^{2^n})a^{2^{n+1] \\
=(a^4-a^8)a^{16} \\
=(a^4-a^5)a^{16}+(a^5-a^6)a^{16}+(a^6-a^7)a^{16}+(a^7-a^8)a^{16} \\
<(a^4-a^5)a^{10}+(a^5-a^6)a^{12}+(a^6-a^7)a^{14}+(a^7-a^8)a^{16} \\
=\sum_{i=2^{n-1]^{2^n-1}(a^i-a^{i+1})a^{2i+2}
\end{split}
\]

When n is summed from n=1 to infinity, then i is also summed from i=1 to infinity.

Title: Re: ***Spoiler*** inequality
Post by James Fingas on Jun 24^th, 2003, 11:24am

Okay, I see where you're coming from now. Too non-obvious for me though...

Title: Re: ***Spoiler*** inequality
Post by towr on Jun 24^th, 2003, 11:29am

for those who'd rather not read latex, even though/if they can/could
http://www.ai.rug.nl/~towr/PHP/FORMULA/formula.php?md5=70bb8146dfb1fa6ee85753467a268f6d