| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Proposed by Francisco Bellot Rosado, I.B. Emilio Ferrari, Valladolid,
Spain.
For every positive integer n, let a[n] be the biggest odd factor of n.
Calculate the sum of the series
a[1] a[2] a[3]
---- + ---- + ---- + ... .
1^3 2^3 3^3
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1798.1 | Pi^2 / 7 | CADSYS::COOPER | Topher Cooper | Mon Sep 20 1993 16:55 | 0 |
| 1798.2 | additional to .1 | AUSSIE::GARSON | nouveau pauvre | Tue Sep 21 1993 19:17 | 43 |
re .0
> For every positive integer n, let a[n] be the biggest odd factor of n.
> Calculate the sum of the series
>
> a[1] a[2] a[3]
> ---- + ---- + ---- + ... .
> 1^3 2^3 3^3
Let n be a positive integer. Then n has a unique representation in the form
2^k(2m+1), k>=0,m>=0. Hence a[n] = 2m+1.
inf
Define S = sigma a[n]/n�
n=1
inf inf 2m+1
Then S = sigma sigma -------------
k=0 m=0 (2^k)�(2m+1)�
inf 1 inf 1
= sigma ------- sigma -------
m=0 (2m+1)� k=0 (2�)^k
8 inf 1
= - � sigma ------- since 1+r+r�+... = 1/(1-r) and putting r=1/8
7 m=0 (2m+1)�
inf
Now it is well known that sigma 1/m� = pi�/6 [EFTR]
m=1
inf 1
hence sigma 1/(2m)� = - � pi�/6 = pi�/24
m=1 4
inf
hence sigma 1/(2m+1)� = pi�/6 - pi�/24 = 3pi�/24 = pi�/8
m=0
8 pi�
Thus S = - � --- = pi�/7 (as claimed in .1)
7 8
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| 1798.3 | generalisation | AUSSIE::GARSON | nouveau pauvre | Thu Sep 23 1993 00:48 | 6 |
Generalising to a divisor other than 2 i.e. let a[n] be the largest
factor of n that is not divisible by d, we get that
d�+d pi�
S[d] = ------ � ---
d�+d+1 6
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