Comments on: Infinite sums, integer sequences http://nealabq.com/blog/2009/03/19/sums_sequences/ Probing dark corners while dodging the grues Thu, 22 Jul 2010 01:02:34 +0000 hourly 1 http://wordpress.org/?v=3.0 By: Peter Luthy http://nealabq.com/blog/2009/03/19/sums_sequences/comment-page-1/#comment-80 Peter Luthy Sat, 18 Apr 2009 08:00:14 +0000 http://nealabq.com/blog/?p=1134#comment-80 Hi Neal, Thanks very much for your kind words, and I'm glad that you found the information on my post interesting. I read your page and am happy to hear that you are teaching your son mathematics and experimental mathematics. The confidence that one can discover things on one's own is a fantastic gift to give to a child. Peter Hi Neal,

Thanks very much for your kind words, and I’m glad that you found the information on my post interesting. I read your page and am happy to hear that you are teaching your son mathematics and experimental mathematics. The confidence that one can discover things on one’s own is a fantastic gift to give to a child.

Peter

]]> By: Neal http://nealabq.com/blog/2009/03/19/sums_sequences/comment-page-1/#comment-79 Neal Fri, 17 Apr 2009 17:34:42 +0000 http://nealabq.com/blog/?p=1134#comment-79 Related post: http://cornellmath.wordpress.com/2009/04/05/a-sill-infinite-series/ Related post:
http://cornellmath.wordpress.com/2009/04/05/a-sill-infinite-series/

]]> By: Neal http://nealabq.com/blog/2009/03/19/sums_sequences/comment-page-1/#comment-72 Neal Wed, 25 Mar 2009 18:24:01 +0000 http://nealabq.com/blog/?p=1134#comment-72 Thanks Zealot. I'd never get that on my own. I guess you've also proven that sum((n^2)/(B^n)) is (B(B+1))/((B-1)^3). You just have to plug in B wherever there's a 10. Anyway, I'll show it to my son. It'll be a good excuse to introduce him to differentials. Thanks Zealot. I’d never get that on my own. I guess you’ve also proven that sum((n^2)/(B^n)) is (B(B+1))/((B-1)^3). You just have to plug in B wherever there’s a 10.

Anyway, I’ll show it to my son. It’ll be a good excuse to introduce him to differentials.

]]> By: Zealot http://nealabq.com/blog/2009/03/19/sums_sequences/comment-page-1/#comment-71 Zealot Tue, 24 Mar 2009 05:44:10 +0000 http://nealabq.com/blog/?p=1134#comment-71 prove of sum ((n^2)/(10^n)) = 110/729: first we have 1/(1-x/10) = sum (x/10)^n then (1/(1-x/10))' = sum ((x/10)^n)' we get (10/(10-x)^2) = sum (n*(x/10)^(n-1)/10) times x to both side (10x/(10-x)^2) = sum (n*(x/10)^n) calculate differential again 10*(1/(10-x)^2 + 2x/(10-x)^3) = sum ((n^2)*(x/10)^(n-1)/10) times x to both side again 10x*(10+x)/(10-x)^3 = sum ((n^2)*(x/10)^n) let x = 1, it is 10*11/9^3 = sum ((n^2)*(x/10)^n), yeah it is 110/729 prove of sum ((n^2)/(10^n)) = 110/729:

first we have 1/(1-x/10) = sum (x/10)^n
then (1/(1-x/10))’ = sum ((x/10)^n)’
we get (10/(10-x)^2) = sum (n*(x/10)^(n-1)/10)
times x to both side (10x/(10-x)^2) = sum (n*(x/10)^n)
calculate differential again 10*(1/(10-x)^2 + 2x/(10-x)^3) = sum ((n^2)*(x/10)^(n-1)/10)
times x to both side again 10x*(10+x)/(10-x)^3 = sum ((n^2)*(x/10)^n)

let x = 1, it is 10*11/9^3 = sum ((n^2)*(x/10)^n), yeah it is 110/729

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