A Nested Radical

Here I present solution to two problems posted on Brilliant: Family of 3's and Infinite sum of reciprocal products and provide a solution to the general expressions.

We know, nnaaa=na1\sqrt[a]{n\sqrt[a]{n\sqrt[a]{\cdots}}}=\sqrt[a-1]{n}

modifying the above one we get,

nn(b1)/(a1)n(c1)/(a1)n(d1)/(a1)edcba=na1\sqrt[a]{n\sqrt[b]{n^{(b-1)/(a-1)}\sqrt[c]{n^{(c-1)/(a-1)}\sqrt[d]{n^{(d-1)/(a-1)}\sqrt[e]{\cdots}}}}}=\sqrt[a-1]{n}

Putting a=x+1,b=2x+1,c=3x+1,d=4x+1,e=5x+1,a=x+1,b=2x+1,c=3x+1,d=4x+1,e=5x+1,\cdots we get,

nn2n3n45x+14x+13x+12x+1x+1=nx\displaystyle \color{#3D99F6} \sqrt[x+1]{n\sqrt[2x+1]{n^2\sqrt[3x+1]{n^3\sqrt[4x+1]{n^4\sqrt[5x+1]{\cdots}}}}}=\sqrt[x]{n}

Putting n=3,x=3n=3, x=3 we get, the nested radical as in Family of 3's :

332 33 a4a3a2a1=33\sqrt[a_1]{3 \sqrt[a_2]{3^2 \ \sqrt[a_3]{3^3 \ \sqrt[a_4]{\cdots}}}}=\sqrt[3]{3} where, an=3n+1a_n=3n+1

Now, nx=nn2n3n45x+14x+13x+12x+1x+1\displaystyle \sqrt[x]{n}=\sqrt[x+1]{n\sqrt[2x+1]{n^2\sqrt[3x+1]{n^3\sqrt[4x+1]{n^4\sqrt[5x+1]{\cdots}}}}}

or,

n1x=(n(n2(n3()14x+1)13x+1)12x+1)1x+1=exp(lnn(1x+1+21x+112x+1+31x+112x+113x+1+))where exp(x)=ex.n^\frac 1{x} = \left(n \left(n^2 \left(n^3 \left(\cdots \right)^\frac 1{4x+1} \right)^\frac 1{3x+1} \right)^\frac 1{2x+1} \right)^\frac 1{x+1} \\ = \exp \left(\ln n {\color{#3D99F6} \left(\frac 1{x+1} + 2 \cdot \frac 1{x+1} \cdot \frac 1{2x+1} + 3 \cdot \frac 1{x+1} \cdot \frac 1{2x+1} \cdot \frac 1{3x+1} + \cdots \right)} \right) \small \color{#3D99F6} \text{where }\exp (x) = e^x.

hence,

1x=(1x+1+21x+112x+1+31x+112x+113x+1+)\displaystyle \color{#3D99F6} \frac 1{x}=\left(\frac 1{x+1} + 2 \cdot \frac 1{x+1} \cdot \frac 1{2x+1} + 3 \cdot \frac 1{x+1} \cdot \frac 1{2x+1} \cdot \frac 1{3x+1} + \cdots \right)

Putting x=3x=3 we get the infinite series in Infinite sum of reciprocal products:

14+24×7+34×7×10+44×7×10×13+=13\displaystyle \frac 14 + \frac 2{4 \times 7} + \frac 3{4 \times 7 \times 10} + \frac 4{4 \times 7 \times 10 \times 13} + \cdots = \frac 1{3}

In fact if ai2a_i \geq 2 and is an integer then,

limsk=1si=1nkt=1k(tai+1)=i=1n1ai\displaystyle \lim_{s \to \infty}\sum_{k = 1}^s \sum_{i=1}^n \frac k{\prod_{t=1}^k(t\cdot a_i +1)}=\sum_{i=1}^n \frac 1{a_i}

Using the same nested radicals as above I have derived the following:

  • n=163nn2n3543=2016\displaystyle \sum_{n=1}^{63} \sqrt{n\sqrt[3]{n^2\sqrt[4]{n^3\sqrt[5]{\cdots}}}}=2016
  • limmj=1m222 23 a4a3a2a1=2π263.127\displaystyle \lim_{m \to \infty} \prod_{j=1}^m \sqrt[a_1]{2 \sqrt[a_2]{2^2 \ \sqrt[a_3]{2^3 \ \sqrt[a_4]{\cdots}}}}=2^\frac{\pi^2}6 \approx 3.127

Thank You\color{#E81990} \text{Thank You}

Note by Mrigank Shekhar Pathak
1 year, 8 months ago

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Wow, I do not understand this at all. Thanks for sharing though.

James Wilson - 1 year, 7 months ago

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Good way but should have should have shown more intermediate steps

Bhavana Bunsha - 1 year, 6 months ago

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Actually there is an easy way to do it, I will be modifying it soon

Mrigank Shekhar Pathak - 1 year, 6 months ago

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