Question

Can someone please help me prove that

√(1+1×√(1+2×√(1+3×√(1+4×√(1+...))))) = 3

Answer 1

Answer:

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Yes

Answer 2

Answer:

Step-by-step explanation:1. 1 + 1 + √1 + ⋯

(a) We put x = 1 + 1 + √1 + ⋯

Then x

 = 1 + 1 + √1 + ⋯

x

 − 1 = 1 + √1 + ⋯ = x

We get a quadratic equation : x

 − x − 1 = 0

Solving, we have x =

 √

 

≈ 1.6180339887 …

Note that the negative root is rejected since x > 0.

You may also notice that 1 + 1 + √1 + ⋯ =

 √

 

= φ, the famous Golden Ratio.

(b) It seems that our job is done, but in fact we still need to show the convergence

of 1 + 1 + √1 + ⋯.

We write x  = √1 = 1

x  = 1 + √1 = √2 ≈ 1.4142

x = 1 + 1 + √1 = 1 + √2 ≈ 1.5538

x = 1 + 1 + √2 ≈ 1.5981

…. x = 1 + 1 + 1 + ⋯ + √1 (n square roots)

We note that: (1) The sequence x ‘may be’ increasing.

(2) x = 1 + x

(i) To prove x is increasing, we use Mathematical Induction.

Let P(n) : x < !  

P(1) is true since x  = 1 < √2 = x  

Assume P(k) is true for some k∈N , that is x" < !"   …(1)

For P(k + 1), From (1) 1 + x" < 1 + !"  

x"   = 1 + x" < 1 + x"   = x"  

∴ P(k + 1) is true.

By the Principle of Mathematical Induction, P(n) is true ∀n∈N .

(ii) To prove x is bounded, we also use Mathematical Induction.

Let P(n) : x < 2 (We use 2 here instead of   √

 

to simplify our writing.)

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