Nested Square roots
Yue Kwok Choy
Nested square roots problems are very interesting. In this article, we investigate some
mathematical techniques applied to this topic that most senior secondary school students can
understand.
1.
√1 + √1 + √1 + ⋯
(a) We put
Then
x = √1 + √1 + √1 + ⋯
x 2 = 1 + √1 + √1 + ⋯
x 2 − 1 = √1 + √1 + ⋯ = x
x2 − x − 1 = 0
We get a quadratic equation :
Solving, we have
x=
1+√5
2
≈ 1.6180339887 …
Note that the negative root is rejected since x > 0.
1+√5
You may also notice that √1 + √1 + √1 + ⋯ =
= φ, the famous Golden Ratio.
2
(b) It seems that our job is done, but in fact we still need to show the convergence
of
√1 + √1 + √1 + ⋯.
We write
x1 = √1 = 1
x2 = √1 + √1 = √2 ≈ 1.4142
x3 = √1 + √1 + √1 = √1 + √2 ≈ 1.5538
x4 = √1 + √1 + √2 ≈ 1.5981
….
We note that:
xn = √1 + √1 + √1 + ⋯ + √1
(1) The sequence
xn
(n square roots)
‘may be’ increasing.
(2) xn = √1 + xn−1
1
We apply the Monotone Convergence Theorem, which states that every monotonic
increasing (or decreasing) sequence bounded above (below) has a limit.
(i)
To prove xn is increasing, we use Mathematical Induction.
Let P(n) :
xn < 𝑥n+1
P(1) is true since
x1 = 1 < √2 = x2
Assume P(k) is true for some kN , that is xk < 𝑥k+1
For P(k + 1), From (1)
1 + xk < 1 + 𝑥k+1
…(1)
xk+1 = √1 + xk < √1 + xk+1 = xk+2
 P(k + 1) is true.
By the Principle of Mathematical Induction,
P(n) is true nN .
(ii) To prove xn is bounded, we also use Mathematical Induction.
Let P(n) :
xn < 2
(We use 2 here instead of
1+√5
2
to simplify our writing.)
x1 = 1 < 2
P(1) is true since
Assume P(k) is true for some kN , that is
For P(k + 1), From (2)
1 + xk < 3
xk < 2
…(2)
xk+1 = √1 + xk < √3 < 2
 P(k + 1) is true.
By the Principle of Mathematical Induction,
P(n) is true nN .
Finally we use the Monotone Convergence Theorem, xn has a limit, and
Quiz
lim
n→∞
xn =
1+√5
2
.
In (1) – (3) below, you may omit the proof of convergence.
(1) Show that
√a + √a + √a + ⋯ = 1+√1+4a
(a > 0)
2
2 +4a
(2) Show that
√a + b√a + b√a + ⋯ = b+√b
(3) Show that
√2 − √2 + √2 − √2 + ⋯ = √5−1
2
2
(4) Find the mistake to prove that
In (1), take limit
But obviously,
a  0,
lim
a→0
(a, b > 0)
1=0:
√a + √a + √a + ⋯ = lim 1+√1+4a = 1
a→0
2
√0 + √0 + √0 + ⋯ = 0
 1=0.
2
2.
√2 + √2 + √2 + ⋯
We change our direction to employ trigonometry to tackle this.
π
cos 4 =
π
√2
2
3.
π
4
√2(1+sin )
cos 8 =
π
(for those who don’t know radian, take π radian = 180o )
2
π
8
=
√2+√2
√2(1+sin )
=
2
(Here we use half-angle formula. Please fill in the calculations.)
√2+√2+√2
….
cos 16 =
2
Hence,
lim
√2 + √2 + √2 + ⋯ = 2 n→∞
cos
2
π
2n
=2
√1 + 2√1 + 3√1 + 4√1 + ⋯
The Indian mathematics genius Ramanujan once published
this problem in the Indian Mathematical journal. He waited for more
than 6 months, but no one came forward with a solution.
He then discovered that:
x+n+a =
√ax + (n + a)2 + x√a(x + n) + (n + a)2 + (x + n)√a(x + 2n) + (n + a)2 + (x + 2n)√…
… (3)
This problem is then a special case where a = 0, n = 1, x = 2.
We don’t discuss the story here as it is a bit involved. Readers interested may study:
http://en.wikipedia.org/wiki/Denesting_radicals#Infinitely_nested_radicals
However, we can still carry out our informal investigation on
√1 + 2√1 + 3√1 + 4√1 + ⋯
Firstly study that,
x1 = √1 = 1
x2 = √1 + 2√1 = √3 ≈ 1.7321
x3 = √1 + 2√1 + 3√1 ≈ 2.2361
3
x4 = √1 + 2√1 + +3√1 + 4√1 ≈ 2.5598
We therefore guess that :
xn = √1 + 2√1 + +3√1 + 4√1 + ⋯ n√1 → 3
We then apply the following identity: n = √1 + (n − 1)(n + 1), for n ≥ 1
repeatedly:
3 = √1 + 2 × 4 = √1 + 2√3 × 5 = √1 + 2√1 + 3√1 + 4 × 6
= ⋯ = √1 + 2√1 + 3√1 + 4√1 + ⋯ + (k − 1)√1 + k(k + 2)
??
= √1 + 2√1 + 3√1 + 4√1 + ⋯
The proof above may already look for most readers, but I still put “??” in the last step, as it
seems to be a not so mathematical “deduction”.
Studying the last expression above more thoroughly, we may guess a more general formula :
4 = √1 + 3√1 + 4√1 + ⋯
5 = √1 + 4√1 + 5√1 + ⋯
…
n = √1 + (n − 1)√1 + n√1 + (n + 1)√1 + ⋯
4
If we assume 3 = √1 + 2√1 + 3√1 + 4√1 + ⋯ is proved, we can begin our induction as
follows:
n = √1 + (n − 1)√1 + n√1 + (n + 1)√1 + ⋯
Let P(n) :
Assume P(k) is true for some kN , that is
k = √1 + (k − 1)√1 + k√1 + (k + 1)√1 + ⋯
For P(k + 1),
From (4),
…(4)
k 2 = 1 + (k − 1)√1 + k√1 + (k + 1)√1 + ⋯
k 2 − 1 = (k − 1)√1 + k√1 + (k + 1)√1 + ⋯
(k − 1)(k + 1) = (k − 1)√1 + k√1 + (k + 1)√1 + ⋯
k + 1 = √1 + k√1 + (k + 1)√1 + ⋯
 P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true nN .
So we have proved n = √1 + (n − 1)√1 + n√1 + (n + 1)√1 + ⋯
which is a step closer
to the Ramanujan formula given by (3).
Quiz
3
Find
√1 + 3√1 + 3√1 + ⋯
5