The Series ( ((((0 +
1)^1 + 1)^1 + 1)^1 + 1)^1 + 1)^1 + … + 1)^1 and Fibonacci Numbers
Add One Then
Reciprocate
Define the series t as:
t = ( ((((0 + 1)^1 + 1)^1 + 1)^1 + 1)^1 + 1)^1 + … +
1)^1 (an infinite amount of terms)
This is a sum that can’t be easily stated in summation
statement (Σ f(x)).
On the HP Prime, I programmed this as:
EXPORT TEST1112(n)
BEGIN
LOCAL k, t:=0;
FOR k FROM 1 TO n DO
t:=(t+1)^1;
END;
RETURN t;
END;
The result seems to converge at 0.6180339785 when n ≥
27. Note that 0.6180339785 = ϕ – 1,
where ϕ is the Golden Ratio ( ϕ = (√5 + 1)/2)
Fibonacci Gets
Involved
Note that:
k =

t =

1

1

2

(1 + 1)^1 = 1/2

3

(1 + 1/2)^1 = (3/2)^1 = 2/3

4

(1 + 2/3)^1 = (5/3)^1 = 3/5

5

(1 + 3/5)^1 = (8/5)^1 = 5/8

6

(1 + 5/8)^1 = (13/8)^1 = 8/13

7

(1 + 8/13)^1 = (21/13)^1 = 13/21

We get a sequence of terms {1, 1/2, 2/3, 3/5, 5/8, 8/13,
13/21, 21/34, 34/55, 55/89, 89/144, …} where each term takes the fraction a/b,
a is the kth Fibonacci number and b is the (k+1)th Fibonacci number. Can we show that this sequence of partial
sums is convergent?
Each partial sums of the series takes the form F_k / F_k+1
where F is the Fibonacci number.
The closed formula for the Fibonacci number is:
F_k = ( ϕ^k – α^k )/√5 , where ϕ = (1 + √5)/2 and α = (1  √5)/2.
Then:
F_k / F_k+1
=( ϕ^k – α^k )/√5 * √5/( ϕ^k+1 – α^k+1)
=( ϕ^k – α^k) / ( ϕ^k+1 – α^k+1 )
= ( ϕ^k / ϕ^k+1) * ( (1 – (α/ϕ)^k) / (1 – (α/ϕ)^k+1) )
= 1/ϕ * ( (1 – (α/ϕ)^k) / (1 – (α/ϕ)^k+1) )
Note that α/ϕ = (1  √5)/(1 + √5 ) ≈ 0.38197 < 1
As k → ∞, α/ϕ →
0.
Hence,
lim k → ∞ (F_k / F_k+1)
= lim k → ∞ (1/ϕ * ( (1 – (α/ϕ)^k) / (1 – (α/ϕ)^k+1) ) )
= 1/ϕ
Simplifying:
1/ϕ
= 2/(1 + √5)
= 2*(1  √5) / ((1 + √5)*(1  √5))
= 2*(1  √5)/4
= (√5 – 1)/2
= √5/2 – 1/2
Adding and subtracting 1/2:
√5/2 – 1/2 + (1/2 – 1/2)
= (√5 + 1)/2 – 1
= ϕ – 1
Since the sequence of partial sums converge to ϕ – 1, the
series
t = ( ((((0 + 1)^1 + 1)^1 + 1)^1 + 1)^1 + 1)^1 + … +
1)^1
converges to ϕ – 1.
This blog is property of Edward Shore. 2015.