Faraz's reciprocal sequence

Algebra Level 5

Consider the sequence { a i } i ≥ 0 \{a_i\}_{i\geq0} such that a 0 = 7 + 1 a_0=\sqrt{7}+1 and a n + 1 = a n 2 − a n + 1 a_{n+1}=a_n^2-a_n+1 . The sum ∑ n = 0 ∞ 1 a n , \sum_{n=0}^{\infty} \frac{1}{a_n}, can be written as 1 b \frac{1}{\sqrt{b}} . What is the value of b b ?

This problem is shared by Faraz M . from AMSP.


The answer is 7.

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Calvin Lin by Calvin Lin

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4 solutions

Anh Tuong Nguyen
Sep 17, 2013

a n + 1 − 1 = a n ( a n − 1 ) a_{n+1}-1=a_n(a_n-1) → 1 a n + 1 − 1 = 1 a n ( a n − 1 ) = 1 a n − 1 − 1 a n \rightarrow \frac{1}{a_{n+1}-1}=\frac{1}{a_n(a_n-1)}=\frac{1}{a_n-1}-\frac{1}{a_n}

→ 1 a n = 1 a n − 1 − 1 a n + 1 − 1 \rightarrow \frac{1}{a_n} = \frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}

∑ n = 0 ∞ 1 a n = lim ⁡ N → ∞ ∑ n = 0 N 1 a n \displaystyle \sum_{n=0}^\infty \frac{1}{a_n}= \lim_{N \to \infty} \displaystyle \sum_{n=0}^N \frac{1}{a_n}

= lim ⁡ N → ∞ ∑ n = 0 N ( 1 a n − 1 − 1 a n + 1 − 1 ) = lim ⁡ N → ∞ ( 1 a 0 − 1 − 1 a N + 1 − 1 ) = \lim_{N \to \infty} \displaystyle \sum_{n=0}^N ( \frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}) = \lim_{N \to \infty} (\frac{1}{a_0-1}-\frac{1}{a_{N+1}-1})

= 1 a 0 − 1 = 1 7 =\frac{1}{a_0-1}=\frac{1}{\sqrt{7}}

16 Helpful 0 Interesting 2 Brilliant 0 Confused

Moderator note:

Yes, thank you! This is the proper way of performing telescoping: doing in for the finite partial sums and then taking a limit.

Jatin Yadav
Sep 17, 2013

First of all , we check that 1 a ∞ = 0 = 1 a ∞ − 1 \frac{1}{a_{\infty}} = 0 = \frac{1}{a_{\infty} - 1}

a n + 1 − 1 = ( a n ) ( a n − 1 ) a_{n + 1} - 1 = (a_{n})( a_{n} - 1 )

Taking reciprocal on both sides ,

1 a n + 1 − 1 = 1 a n − 1 − 1 a n \frac{1}{a_{n + 1} - 1} = \frac{1}{a_{n} - 1} - \frac{1}{a_{n}}

⇒ 1 a n = 1 a n − 1 − 1 a n + 1 − 1 \Rightarrow \frac{1}{a_{n}} = \frac{1}{a_{n} - 1} - \frac{1}{a_{n + 1} - 1}

∑ 1 = 0 ∞ 1 a i = ∑ i = 0 ∞ 1 a i − 1 − 1 a i + 1 − 1 \displaystyle \sum_{1 = 0}^{\infty} \frac{1}{a_{i}} = \sum_{i = 0}^{\infty} \frac{1}{a_{i} - 1} - \frac{1}{a_{i + 1} - 1}

= 1 a 0 − 1 − 1 a ∞ − 1 \frac{1}{a_{0} - 1} - \frac{1}{a_{\infty} - 1}

= 1 7 \frac{1}{\sqrt{7}} ⇒ b = 7 \Rightarrow b = \fbox{7}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

It is in general not a good idea to operate with a ∞ . a_{\infty}. It is not really defined. You can define it in this case as infinity, but only because lim ⁡ i → ∞ a i = + ∞ , \lim \limits_{i\to \infty} a_i=+\infty, which has to be proven. And still the most rigorous(if not the only rigorous) way to proceed is to do the telescoping for the finite sums and then take the limit.

Alexander Borisov - 7 years, 8 months ago
Obwj Obid
Sep 16, 2013

We shall prove it in general that 1 a 0 − 1 = ∑ k = 0 ∞ 1 a k \frac{1}{a_{0}-1}=\sum_{k=0}^{\infty}\frac{1}{a_k} .First, note that the identity 1 a k − 1 − 1 a k = 1 ( a k ) 2 − a k \frac{1}{a_{k}-1}-\frac{1}{a_{k}}=\frac{1}{(a_k)^2-a_k} or 1 a k − 1 = 1 a k + 1 a k + 1 − 1 \frac{1}{a_{k}-1}=\frac{1}{a_{k}}+\frac{1}{a_{k+1}-1} .then 1 a 0 − 1 = 1 a 0 + 1 a 1 − 1 = 1 a 0 + 1 a 1 + 1 a 2 − 1 = . . . . . . = ∑ k = 0 ∞ 1 a k \frac{1}{a_{0}-1}=\frac{1}{a_{0}}+\frac{1}{a_{1}-1}=\frac{1}{a_{0}}+\frac{1}{a_{1}}+\frac{1}{a_{2}-1}=......=\sum_{k=0}^{\infty}\frac{1}{a_k} .So, let a 0 = 7 + 1 a_0=\sqrt{7}+1 ,then b = ( 7 + 1 ) − 1 , b = 7 \sqrt{b}=(\sqrt{7}+1)-1,b=7

4 Helpful 0 Interesting 0 Brilliant 0 Confused

One has to be careful: this only works because lim ⁡ i → ∞ 1 a i − 1 = 0 \lim \limits_{i\to \infty} \frac{1}{a_i-1}=0

Alexander Borisov - 7 years, 8 months ago
Zi Song Yeoh
Sep 16, 2013

Firstly, note that 1 a n = 1 a n − 1 − 1 a n + 1 − 1 \frac{1}{a_n} = \frac{1}{a_n - 1} - \frac{1}{a_{n + 1} - 1} . Next, ∑ i = 0 ∞ 1 a n = ∑ i = 0 ∞ ( 1 a n − 1 + 1 a n − 1 − 1 ) = 1 a 0 − 1 = 1 7 \sum^{\infty}_{i = 0}\frac{1}{a_n} = \sum^{\infty}_{i = 0}(\frac{1}{a_n - 1} + \frac{1}{a_{n - 1} - 1}) = \frac{1}{a_0 - 1} = \frac{1}{\sqrt{7}} . So, b = 7 b = \boxed{7} .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

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