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, $\sqrt[a]{n\sqrt[a]{n\sqrt[a]{\cdots}}}=\sqrt[a-1]{n}$

modifying the above one we get,

$\sqrt[a]{n\sqrt[b]{n^{(b-1)/(a-1)}\sqrt[c]{n^{(c-1)/(a-1)}\sqrt[d]{n^{(d-1)/(a-1)}\sqrt[e]{\cdots}}}}}=\sqrt[a-1]{n}$

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

$\displaystyle \color{#3D99F6} \sqrt[x+1]{n\sqrt[2x+1]{n^2\sqrt[3x+1]{n^3\sqrt[4x+1]{n^4\sqrt[5x+1]{\cdots}}}}}=\sqrt[x]{n}$

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

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

Now, $\displaystyle \sqrt[x]{n}=\sqrt[x+1]{n\sqrt[2x+1]{n^2\sqrt[3x+1]{n^3\sqrt[4x+1]{n^4\sqrt[5x+1]{\cdots}}}}}$

or,

$n^\frac 1{x} = \left(n \left(n^2 \left(n^3 \left(\cdots \right)^\frac 1{4x+1} \right)^\frac 1{3x+1} \right)^\frac 1{2x+1} \right)^\frac 1{x+1} \\ = \exp \left(\ln n {\color{#3D99F6} \left(\frac 1{x+1} + 2 \cdot \frac 1{x+1} \cdot \frac 1{2x+1} + 3 \cdot \frac 1{x+1} \cdot \frac 1{2x+1} \cdot \frac 1{3x+1} + \cdots \right)} \right) \small \color{#3D99F6} \text{where }\exp (x) = e^x.$

hence,

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

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

$\displaystyle \frac 14 + \frac 2{4 \times 7} + \frac 3{4 \times 7 \times 10} + \frac 4{4 \times 7 \times 10 \times 13} + \cdots = \frac 1{3}$

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

$\displaystyle \lim_{s \to \infty}\sum_{k = 1}^s \sum_{i=1}^n \frac k{\prod_{t=1}^k(t\cdot a_i +1)}=\sum_{i=1}^n \frac 1{a_i}$

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

• $\displaystyle \sum_{n=1}^{63} \sqrt{n\sqrt[3]{n^2\sqrt[4]{n^3\sqrt[5]{\cdots}}}}=2016$
• $\displaystyle \lim_{m \to \infty} \prod_{j=1}^m \sqrt[a_1]{2 \sqrt[a_2]{2^2 \ \sqrt[a_3]{2^3 \ \sqrt[a_4]{\cdots}}}}=2^\frac{\pi^2}6 \approx 3.127$

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

Note by Mrigank Shekhar Pathak
3 years, 4 months ago

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

- 3 years, 4 months ago

Good way but should have should have shown more intermediate steps

- 3 years, 2 months ago

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

- 3 years, 2 months ago

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- 1 year, 9 months ago