Question

Find the value of the following -
$\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{... \infty } } } } } } }$
​

Answer 1

We have a Nested square values with a value 2 in each square root up to infinity times. We have to expand expand 2 to 3 terms and we have to check the pattern for which the nested square roots tends up to infinity. We have the nested square roots :

$\sf{\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{\ldots \infty}}}}}}}}$

First we need to identify the series of exponents that will occur when we reduce square roots one by one. For that we have to check some 2 to 3 terms of the given nested square roots taking as radical.

$\longrightarrow\textsf{We can write \bf{\displaystyle \sqrt{2}=(2)^{\frac{1}{2}}} using radical form}$

Consider first two nested roots from the whole nested roots of 2

$:\implies\sf{\displaystyle\sqrt{2 \sqrt{2}}=\left(2 \times 2^{\frac{1}{2}}\right)^{\frac{1}{2}}}\sf{\implies \sqrt{2 \sqrt{2}}=2^{\frac{1}{2}+\frac{1}{4}}}\quad(\textbf{ \bf{\because} \bf{a^m\times a^n=a^{m+n}}) }$

If we consider three terms from the above nested roots of 2

$:\implies\sf{\displaystyle\sqrt{2 \sqrt{2 \sqrt{2}}}=\left(2 \times\left(2(2)^{\frac{1}{2}}\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}}\implies\sf{\displaystyle\sqrt{2 \sqrt{2 \sqrt{2}}}=2 ^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}}$

$\dashrightarrow\textsf{Hence if we take \tiny{\textsf{ \sf{\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{\ldots \infty}}}}}}}} }} then we get \sf{2 ^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}} }$

Thus we get a series as exponent for 2. Now we have to find the sum the series.  Now consider the series $\sf{^1/_2+\:^1/_4+\:^1/_8+\ldots \ldots}$ The above series satisfies the properties of geometric series. We have geometric series with initial value $\sf{a=\:}\sf{^1/_2}$ and common ratio (term 2 / term 1)

$:\implies\sf{\displaystyle r=\frac{t_{2}}{t_{1}}=1/4\:\div\:1/2=2/4=1/2 }$

We have r < 1, sum of the geometric series is calculated by the formula $\boxed{\bf{\mathbf{S}=\frac{\mathbf{F} \textbf{irst } \mathbf{T} \textbf{erm }}{1-\textbf{ Common } \textbf{Ratio }}}}$

$:\implies\displaystyle\sf{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \ldots \ldots=\frac{\frac{1}{2}}{1-\frac{1}{2}}}\implies\displaystyle\sf{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \ldots \ldots=1}$

• $\textsf{Now substitute sum of the series to \displaystyle\sf{2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots}} , \bf{We\: get\: 2^1 = 2} }$

Hence we got the required solution

Answer 2

Given : $\sf{\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{\ldots \infty}}}}}}}}$

To find : Value

Solution:

x =$\sf{\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{\ldots \infty}}}}}}}}$

squaring both sides

$x^2 = 2 \sf{\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{\ldots \infty}}}}}}}}$

$x^2 = 2x$

dividing by x on both sides

x = 2

$\sf{\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{\ldots \infty}}}}}}}} = 2$