Find the limit of the sequence

{√(2), √(2√(2)), √(2√(2√(2))),...

if you consider the "last" term as

x = √(2√(2√(2√(2....)))
then
x^2 = 2√(2√(2√(2√...))) = 2x
x = 2

To find the limit of the given sequence, let's start by analyzing its pattern.

We have a recursive formula for the sequence, where each term is obtained by taking the square root of the previous term and multiplying it by √2:

a₁ = √2
a₂ = √(2 * √2) = √2 * √(√2) = √2 * (2^(1/4)) = 2^(1/4)

We can observe that as we go from each term to the next, the expression inside the square root becomes longer. So, let's express the general term of this sequence:

aₙ = √(2√(2√(...√2)...))

where the expression inside the square root has n square root symbols.

Now, let's see if we can find a pattern for the general term:

a₃ = √(2√(2√2)) = √(2 * 2^(1/4)) = √(2^(5/4)) = 2^(5/8)
a₄ = √(2√(2√(2√2))) = √(2√(2 * 2^(1/4))) = √(2 * 2^(3/2)) = √(2^(5/2)) = 2^(5/4)
...

We can observe that each term can be written as 2 raised to the power of 5 divided by 2^n−1. Therefore, the general term can be expressed as:

aₙ = 2^(5/2^(n−1))

Now, let's find the limit of this sequence as n approaches infinity:

lim(n→∞) aₙ = lim(n→∞) 2^(5/2^(n−1))

As n approaches infinity, 2^(n−1) in the denominator also approaches infinity. Since the exponent of 2 is getting larger and larger, the limit of the expression is 1. Therefore,

lim(n→∞) aₙ = lim(n→∞) 2^(5/2^(n−1)) = 2^0 = 1

Hence, the limit of the given sequence is 1.

To find the limit of the given sequence, let's examine the pattern:

The first term is √2.

The second term is √(2√2) = √(2 * √2) = √(2^(3/2)) = 2^(3/4).

The third term is √(2√(2√2)) = √(2√(2 * √2)) = √(2 * 2^(3/4)) = √(2^(5/2)) = 2^(5/8).

From this pattern, we observe that each term can be written as 2 raised to a power that follows a specific pattern. The power starts with 1/2 (√2 = 2^(1/2)), and in each subsequent term, it increases by multiplying the previous power by 3/4.

Therefore, we can write the nth term of the sequence as 2 raised to the power of (1/2) * (3/4)^(n-1).

To find the limit as n approaches infinity, we evaluate the expression as n becomes larger and larger.

Limit as n approaches infinity of 2^((1/2) * (3/4)^(n-1))

Since (3/4)^(n-1) approaches zero as n approaches infinity, the exponent of 2 becomes (1/2) * 0 = 0.

Thus, we have:

Limit as n approaches infinity of 2^((1/2) * (3/4)^(n-1)) = 2^0 = 1.

Therefore, the limit of the given sequence is 1.