# Evaluate the expression $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdots}}}$

Is there a way to check whether the expression below converges to a specific number. $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdots}}}$$ or in other words does the sequence defined by $x_0=\sqrt{2}$ and $x_n=\sqrt{2}^{x_{n-1}}$ converges?

Trying with a calculator to evaluate an eight-length $\sqrt{2}$ construct I got $1.9656648865173187$ but I couldn't yet confirm convergence.

• – Lucian Sep 21 '15 at 20:38

That sequence converges to exactly $2$. To prove this, consider the function $$f(x)=\sqrt{2}^x$$ it should be reasonably clear that, if $1,f(1),f(f(1)),f(f(f(1))),\ldots$ has a limit $x$, then that limit satisfies $x=f(x)$, since $f$ is continuous. By inspection, one can see that $f(2)=2$ and $f(4)=4$ and that these are the only two real values satisfying this since $f$ is convex. Moreover, one may note that, for any $x<2$ we have $x<f(x)<2$. Putting this together, we can find that $1,f(1),f(f(1)),\ldots$ is an increasing sequence, bounded above by $2$ and hence has a limit. Its limit cannot exceed $2$ but satisfies $f(x)=x$. Therefore, its limit is $2$.