Find the pattern: rationalising the irrational.

Question

The series is:

1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, ...

a) What is the next term?

b) What is the formula for the nth term as a function of integer n, where n>0?

c) This is an infinite series. What value does the nth term converge to?

Answer 1

a) Next number is 1393/985

b)The formula is

let numerator be denoted by n(k) where k belongs to N
and denominator be denoted by d(k) where k belongs to N

for numerator: 2*(n(k-1)) + (n(k-2))

for denominator: 2*(d(k-1)) + (d(k-2))

So the series formula is

(2*(n(k-1)) + (n(k-2)))/(2*(d(k-1)) + (d(k-2)))

c)using limits tends to infinity on the above formula gives the answer 1,

but that is wrong, so I need little more time.

 Edit:

239/169 = 1.41420118343

577/408 = 1.41421568627

1393/985 =1.41421319797,

So I guess nth term is sqrt(2)

Ok???

Comments

One more thing I have observed,

numerator - denominator = previous denominator

also, previous numerator + previous denominator = next denominator

so, maybe I can get a single formula instead of separate formula for numerator and denominator.

Thank You

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Refined formula

n(k) = 2*(n(k-1)) + (n(k-2))

d(k) = n(k-1) + d(k-1)

but d(k-1) = n(k-2) + d(k-2)

so formula of next number can be given as

2*(n(k-1)) + (n(k-2))/ g

where g = n(k-1) + n(k-2) + .................... 1

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Yes, you got it! Can you prove that progressive terms converge to √2?

As for the formula (the hardest part of this question I think), here’s a clue:

Square each term and note the relationship between the denominator and the numerator.

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To find a reliable formula, I did some research and found a clue that pointed me to a solution, which I arrived at after some experimentation through an educated guess. I’m putting together a complete solution which includes deriving the formula as well as demonstrating it.

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Please no, I also did some research and found a way.

Please let me do it, and if I fail then you can continue.

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Yes, of course, I wouldn’t spoil your efforts to find a formula. It will be interesting to compare your findings with mine, when you eventually find a solution. You often do find a solution method which differs from mine, so I will be interested in seeing what you come up with.

Sometimes research leads to solving other problems or finding new methods.

Good luck with your efforts. And best wishes. Also, thank you for your words of appreciation.

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The required formula for finding the nth term must involve n only, so, following your method of finding formulas for the numerator and denominator separately, each formula would be in terms of n only (that is, not involving any previous terms). Here’s a clue (hopefully without spoiling your attempts): There is only really one formula used for both numerator and denominator. But first you need to make a simple change to each term of the given series. This change will reduce the series to a single series from which the numerator and denominator can be derived.