Equation with limit

$\lim\limits_{n\to \infty}\sqrt{1+\sqrt{x+\sqrt{x^2+...+\sqrt{x^n}}}}=2$

I had never faced with problems like this. Give me, please, a little hint.

Solution 1:

Hint. $$f(x) = \sqrt{x+\sqrt{x^2+\sqrt{x^3+\sqrt{x^4+\ldots}}}} $$ is a continuous and increasing function on $\mathbb{R}^+$, as well as: $$ f_k(x) = \sqrt{x^k+\sqrt{x^{k+1}+\sqrt{x^{k+2}+\ldots}}}$$ It is enough to check that $f(4)=3$, or: $$ f_1(4) = 2\cdot 4^0+1, $$ or: $$ f_2(4) = f_1(4)^2-4 = (2\cdot 4^0+1)^2-2^2=4+1 $$ or: $$ f_3(4)=f_2(4)^2-4^2 = (4+1)^2-4^2 = 2\cdot 4+1 $$ or: $$ f_4(4)=f_3(4)^2-4^3 = (2\cdot 4+1)^2 - 4^3 = 4^2+1 $$ or: $$ f_5(4)=f_4(4)^2-4^4 = (4^2+1)^2-(4^2)^2 = 2\cdot 4^2+1 $$ $\ldots$ A pattern emerged. Notice that

$$ f_{2n+1}(4) = 2\cdot 4^n+1 \qquad f_{2n}(4) = 4^n+1 \tag{1}$$

are equivalent, since $$ f_{2n+2}(x)=f_{2n+1}(x)^2-x^{2n+1}, \qquad f_{2n+1}(x)=f_{2n}(x)^2-x^{2n}. $$

So it is enough to prove that

$$ \lim_{n\to +\infty}\frac{f_{2n}(4)}{4^n+1}=\lim_{n\to +\infty}\frac{f_{2n+1}(4)}{2\cdot 4^n+1}=1 \tag{2}$$

hold by squeezing.

Solution 2:

Well, we have to start somewhere, so we might as well guess and check. I made the following Python program:

from math import sqrt
def find_answer(x, n):
    '''Calculates the expression in the limit for given x and n'''
    answer = 0
    for power in range(n, -1, -1):
        answer += x**power
        answer = sqrt(answer)
    return answer

# Here, I guess x from x=2 to x=10 and use n=100 so I am close to the actual limit
for x in range(2, 10):
    print(x, find_answer(x, 100))

$x=4$ gives me an answer of $2.0$, so it seems that the answer to this problem is $x=4$, or at least an $x$ very close to $4$.