username53983805
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#1
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#1
Evaluate the sum S=\frac{1}{1}+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{5}{16}+\frac{8}{32}+\frac{13}{64}+\cdots

I've tried rewriting the sum by splitting terms so that S=(\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\cdots)+(\frac{1}{4}+\frac{2}{8}+\frac{4}{16}+\frac{7}{32}+\frac{12}{64}+\cdots), as the first sum just converges to 2, but I can't seem to really do much else with the second sum.

Is there a better way that I could split this sum to make the problem more manageable?
Last edited by username53983805; 1 month ago
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mqb2766
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#2
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#2
(Original post by username53983805)
Evaluate the sum S=\frac{1}{1}+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{5}{16}+\frac{8}{32}+\frac{13}{64}+\cdots

I've tried rewriting the sum by splitting terms so that S=(\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\cdots)+(\frac{1}{4}+\frac{2}{8}+\frac{4}{16}+\frac{7}{32}+\frac{12}{64}+\cdots), as the first sum just converges to 2, but I can't seem to really do much else with the second sum.

Is there a better way that I could split this sum to make the problem more manageable?
What do you notice about the numerator and denominator of the original sequence?
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username53983805
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#3
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(Original post by mqb2766)
What do you notice about the numerator and denominator of the original sequence?
Numerator is fibonacci, denominator is increasing powers of two
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mqb2766
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#4
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(Original post by username53983805)
Numerator is fibonacci, denominator is increasing powers of two
So think how you can use that info.

As a hint, you can "prove" the standard result of
S = 1 + 1/2 + 1/4 + 1/8 + ... = 2
by noting
S = 1 + S/2
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username53983805
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#5
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(Original post by mqb2766)
So think how you can use that info.

As a hint, you can "prove" the standard result of
S = 1 + 1/2 + 1/4 + 1/8 + ... = 2
by noting
S = 1 + S/2
I now have something.

So \frac{1}{1}+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{5}{16}+\frac{8}{32}+\frac{13}{64}+\cdots=\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{2}{8}+\frac{1}{8}+\frac{3}{16}+\frac{2}{16}+\frac{5}{32}+\frac{3}{32}+\frac{8}{64}+\frac{5}{64}+\cdots
\implies S=1+(\frac{1}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{5}{32}+\frac{8}{64}+\cdots)+(\frac{1}{4}+\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{5}{64}+\cdots)=1+\frac{1}{2}S+\frac{1}{4}S
\implies S=4.
Last edited by username53983805; 1 month ago
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mqb2766
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#6
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#6
Looks about right.
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