Before writing this post, I just want you to know that I'm still a student and my English isn't very good, so it might contain errors. :)--XmL
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According to my "more Interesting stuff" post. I mentioned that:
1/6+1/12+1/20
=1/(2*3)+1/(3*4)+1/(4*5)
=1/2-1/3+1/3-1/4+1/4-1/5
=1/2-1/5
=3/10
Now, let me ask you this: what's 1/6+1/12+1/20+1/30+1/42+............... to infinity?
well, let me start by breaking this down.
We know that 1/6=1/2*3=1/2-1/3, 1/12=1/3*4=1/3-1/4 and so on
so when I put this up: 1/2-1/3+1/3-1/4+1/4-1/5.......................
since -1/n+1/n=0
so 1/6+1/12+1/20+1/30+1/42+..............+ to infinity=1/2 (infinite sum!!!!)
this sum is in a form of 1/a*b+1/b*c+1/c*d+1/d*e+...............+ to infinity=1/a
is it true?
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Now let's explore the form (a/b)^n+(a/b)^(n+1)+(a/b)^(n+2)+..................+to infinity (where b>a)
Solve for S, S=(3/4)^2+(3/4)^3+(3/4)^4+(3/4)^5+................+to infinity --------------(1)
multiply each side by 3/4:
3S/4=(3/4)^3+(3/4)^4+(3/4)^5+................+to infinity -------------------(2)
(1)-(2):
S-3S/4=(3/4)^2
S/4=9/16
S=9/4
S-a*S/b=(a/b)^n
S*(b-a)/b=(a/b)^n
S=[b(a/b)^n]/(b-a)
For this reason,the solution of S=(a/b)^n+(a/b)^(n+1)+(a/b)^(n+2)+..................+to infinity (where b>a)
is S=[b(a/b)^n]/(b-a).
is it true?
