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, 11 months ago

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

James Wilson - 1 year, 11 months ago

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Good way but should have should have shown more intermediate steps

Bhavana Bunsha - 1 year, 10 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, 10 months ago

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Amazing! You done amazing work in this post. The global cash prize for the science students is all time amazing and you're offering very well in this post and now you can write my dissertation to manage your task. Keep it up!

Christian Farmer - 4 months, 4 weeks ago

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