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        riddles >> putnam exam (pure math) >> Summation of series
        
(Message started by: Optimus Prime on Jan 21^st, 2008, 8:19pm)

Title: Summation of series
Post by Optimus Prime on Jan 21^st, 2008, 8:19pm

Find the sum of the first n terms of the following series:
1/7 + 1/77 + 1/777 + ... up to n-th term

Title: Re: Summation of series
Post by ThudanBlunder on Jan 21^st, 2008, 10:23pm

So we need to find
n
(9/7)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif1/(10^k - 1)
k=1

Title: Re: Summation of series
Post by Icarus on Jan 24^th, 2008, 5:13pm

If Optimus Prime has a nice answer for this, it would be interesting from a number theory point-of-view.

If we let x = 0.1, we can rewrite T&B's summation as
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif_k=1..n x^k/(1 - x^k) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif_k=1..n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif_{j=1..http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gif} x^kj = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif_{m=1..http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gif} d(m; n) x^m,

where d(m; n) = |{ d http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbn.gif : d | m and d <= n}| is the number of divisors of m less than or equal to n.

Thus a generalized solution to this problem could provided a means of generating the values of d(m; n).