Question

What is the value of $S = \sqrt {6 + 2\sqrt {7 + 3\sqrt {8 + 4\sqrt {9 + ...} } } }$ ?

Answer 1

In the above question, we are given as infinite series $S = \sqrt {6 + 2\sqrt {7 + 3\sqrt {8 + 4\sqrt {9 + ...} } } }$ . We have to find the value of S. In order to approach the solution, first we have to write the series in its general form i.e. with the help of replacing the numbers with a proper variable. After that we can write the series in the form of a function of $x$ . Now we can obtain $f\left( 0 \right)$ and then similarly $f\left( x \right)$ .

Complete step-by-step answer:
Given series is,
$\Rightarrow S = \sqrt {6 + 2\sqrt {7 + 3\sqrt {8 + 4\sqrt {9 + ...} } } }$
We have to find the value of S.
We can write the above series as a function of $x$ as,
$\Rightarrow f\left( x \right) = \sqrt {x + 4 + x\sqrt {x + 5 + \left( {x + 1} \right)\sqrt {x + 6 + \left( {x + 2} \right)\sqrt {x + 7 + ...} } } }$
Notice here that if we put $x = 2$ then we have,
$\Rightarrow f\left( 2 \right) = \sqrt {6 + 2\sqrt {7 + 3\sqrt {8 + 4\sqrt {9 + ...} } } }$
Therefore,
$\Rightarrow S = f\left( 2 \right)$
So our aim here is to find the value of $f\left( x \right)$ for $x = 2$ i.e. $f\left( 2 \right)$ .
Now we can present a simple method to find the form of $f\left( x \right)$ .
Since $f\left( x \right)$ is always positive and increasing, so we know that,
$\Rightarrow f\left( x \right) < f\left( {x + 1} \right) < f\left( {x + 2} \right) < ... < f\left( {x + n} \right)$
Also we if we put $x = 0$ then we have,
$\Rightarrow f\left( 0 \right) = \sqrt 4$
Or,
$\Rightarrow f\left( 0 \right) = 2$
So at this point, we can write
$\Rightarrow f\left( x \right) = {c_0} + {c_1}x$ ...(1)
Substituting above value into the general form of the series, we have
$\Rightarrow f\left( x \right) = \sqrt {x + 4 + xf\left( {x + 1} \right)}$ ...(2)
From equation (1) and (2) we can write,
$\Rightarrow {c_0} + {c_1}x = \sqrt {x + 4 + xf\left( {x + 1} \right)}$
Squaring both sides, we have
$\Rightarrow {\left( {{c_0} + {c_1}x} \right)^2} = x + 4 + xf\left( {x + 1} \right)$
Now from (1), we can put $f\left( {x + 1} \right) = {c_0} + {c_1}\left( {x + 1} \right)$ , so we get
\Rightarrow {\left( {{c_0} + {c_1}x} \right)^2} = x + 4 + x\left\\{ {{c_0} + {c_1}\left( {x + 1} \right)} \right\\}
After grouping the coefficients, we have the equation as
$\Rightarrow {x^2}\left( {{c_1}^2 - {c_1}} \right) + \left( {2{c_0}{c_1} - 1 - {c_0} - {c_1}} \right)x + \left( {{c_0}^2 - 4} \right) = 0$
Solving the equation, we get the coefficients equal to zero as,
${c_1}^2 - {c_1} = 0$
$2{c_0}{c_1} - 1 - {c_0} - {c_1} = 0$
${c_0}^2 - 4 = 0$
That gives,
$\Rightarrow {c_0} = 2$
And
$\Rightarrow {c_1} = 1$
Putting these values in equation (1), we get
$\Rightarrow f\left( x \right) = 2 + 1 \cdot x$
Our aim was to find $f\left( 2 \right)$ hence putting $x = 2$ in above equation gives,
$\Rightarrow f\left( 2 \right) = 2 + 2$
Hence,
$\Rightarrow f\left( 2 \right) = 4$
That is the required solution.
Therefore, the value of $S = \sqrt {6 + 2\sqrt {7 + 3\sqrt {8 + 4\sqrt {9 + ...} } } }$ is $4$ .

Note: The standard definition of an “ ${n^{th}}$ root” of a number $x$ is another number $y$ such that ${y^n} = x,y \geqslant 0$ . Then $y$ is represented as $\sqrt[n]{x} = {x^{\dfrac{1}{n}}}$ . For getting the “infinity”th root of a number, we take the limit of $\dfrac{1}{n}$ going to infinity.
Now since $\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n} = 0$
Therefore,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } {x^{\dfrac{1}{n}}} = {x^0}$
Or,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } {x^{\dfrac{1}{n}}} = 1$
Thus, we have that $\sqrt[\infty ]{x} = 1$ .