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Brøksum
Posted: 22/02-2010 23:02
by Karl_Erik
Gitt et heltall [tex]n[/tex], finn summen av alle brøker på formen [tex]\frac 1 {ab}[/tex] der [tex]a[/tex] og [tex]b[/tex] er innbyrdes primiske positive heltall, høyst lik [tex]n[/tex], og slik at [tex]a+b>n[/tex].
Posted: 23/02-2010 16:45
by Charlatan
La
[tex] \gamma(n)=\sum_{\tiny{a+b \geq n+1 \\ (a,b)=1 \\ a,b \leq n}} \frac{1}{ab}[/tex]
Da er
[tex] \gamma(n+1) = \sum_{\tiny{a+b \geq n+2 \\ (a,b)=1 \\ a,b \leq n+1}} \frac{1}{ab} = \sum_{\tiny{a+b \geq n+2 \\ (a,b)=1 \\ a,b \leq n}} \frac{1}{ab} + \sum_{\tiny{a=n+1 \\ \text{eller} \\ b=n+1 \\ a,b \leq n+1}} \frac{1}{ab}= \sum_{\tiny{a+b \geq n+1 \\ (a,b)=1 \\ a,b \leq n}} \frac{1}{ab} +\sum_{\tiny{a=n+1 \\ \text{eller} \\ b=n+1 \\ a,b \leq n+1}} \frac{1}{ab} - \sum_{\tiny{a+b = n+1 \\ (a,b)=1 \\ a,b \leq n}} \frac{1}{ab} = \gamma(n)+\sum^{n+1}_{\tiny{a=1 \\ (a,n+1)=1}} \frac{2}{a(n+1)} - \sum^n_{\tiny{a = 1 \\ (a,n+1-a)=1}} \frac{1}{a(n+1-a)} \\ = \gamma(n)+\sum^n_{\tiny{a=1 \\ (a,n+1)=1}} \frac{2}{a(n+1)} - \sum^n_{\tiny{a = 1 \\ (a,n+1)=1}} \frac{1}{a(n+1-a)} = \gamma(n)+\sum^n_{\tiny{a=1 \\ (a,n+1)=1}} \frac{2}{a(n+1)} -\frac{1}{a(n+1-a)} = \gamma(n)+\sum^n_{\tiny{a=1 \\ (a,n+1)=1}} \frac{n+1-a}{a(n+1)(n+1-a)} - \frac{a}{a(n+1)(n+1-a)} [/tex]
Siden [tex](a,n+1)=1 \Leftrightarrow (n+1-a,n+1)=1[/tex] har vi av symmetri i den siste summen at [tex]\gamma(n+1)=\gamma(n)[/tex].
Siden [tex]\gamma(1) = 1[/tex], må [tex]\gamma(n)=1[/tex] for alle [tex]n[/tex].
Summen vi er ute etter er [tex]\frac12 \gamma(n) = \frac12[/tex], siden vi teller hver brøk to ganger i [tex]\gamma(n)[/tex].