Question

Find the sum of the following series:
1/1+1/2+2/4+3/8+5/16+8/32+...

where each numerator is a term of the Fibonacci Sequence and the denominators are in a geometrical progression.

Answer 1

I understand that tn=((tn-1)+(tn-2))/2

Answer 2

You mean (n-1)^2?

Answer 3

denominator is not n^2.....it is 2^(
n-1)

Answer 4

\(\frac11+\frac12+\frac24+\frac38+\frac5{16}+\frac8{32}+\cdots\)

Answer 5

no..its a converging series

Answer 6

sorry ..

Answer 7

oops, converges at 4. my bad. :s

Answer 8

i think phi must be used to solve this problem

Answer 9

\(t_n=\frac {F_n}{2^n}\), where \(F_n\) - n-th term of Fibonacci sequence

Answer 10

@chaise...denominator: 2^(n-1)

Answer 11

n=0,1,2,...

Answer 12

Such confusion! People it's not that hard...

\(\large F_n = F_{n-1} + F_{n-2}\), where n = discrete interger numbers from 1 to \(\infty\)

The n-1 mean "the previous term"
The n-2 means "the one before the previous term"

This is recursion in action here

Answer 13

Right, so how do you determine the sum to infinity?

Answer 14

In this case we need to modify that series so it matches the output above

Answer 15

You need to do a test of some kind @chaise

Alternating series test? no.
P-series? no.
Geometric series? no.
Integral test? no.
Ratio test? ;-) hmm...

Answer 16

http://mathworld.wolfram.com/FibonacciNumber.html
chek out (15)

Answer 17

is it a harmonic progression ?

Answer 18

i do think like that : this is in a harmonic progression ..

Answer 19

put the value of x=1/2
divide the left by x again ... i guess this would give the answer.

Answer 20

@satya_Balli I would suspect it is NOT a convergent series. Even if it's a positive number and the positive numbers go to zero they keep increasing the total sum little by little, just as the area under the following would be infinite:

\[\large \int\limits_{1}^{\infty} \log |x| \ \ dx\]