<s>[INST] Evaluating the limit of $\sqrt[2]{2+\sqrt[3]{2+\sqrt[4]{2+\cdots+\sqrt[n]{2}}}}$ when $n\to\infty$ The following nested radical $$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$$ is known to converge to $2$. 
We can consider a similar nested radical where the degree of the radicals increases: 
$$\sqrt[2]{2+\sqrt[3]{2+\sqrt[4]{2+\cdots+\sqrt[n]{2}}}}$$ 
which converges to a constant $C=1.8695973...$. Is there a closed-form expression for this limit? 
 [/INST] As already mentioned by others, there is no closed form known for this expression (like the majority of infinite nested radicals).
I would consider some bounds for this number, which give quite good approximations. 
First, instead of truncating the number, it's much better to replace the remainder by $1$ at any step, for example:
$$\sqrt{2+\sqrt[3]{2+\sqrt[4]{2+\cdots}}}>\sqrt{2+\sqrt[3]{2+\sqrt[4]{2+1}}}=1.8684804$$
For the upper bound we can use the approximation:
$$\sqrt{2+\sqrt[3]{2+\sqrt[4]{2+\cdots}}}<\sqrt{2+\sqrt[3]{2+\sqrt[3]{2+\cdots}}}$$
To find $\sqrt[3]{2+\sqrt[3]{2+\cdots}}$, we use the equation:
$$x^3-x-2=0$$
$$x=\frac{1}{3} \left(\sqrt[3]{27-3\sqrt{78}}+\sqrt[3]{27+3\sqrt{78}} \right)=1.5213797$$
$$\sqrt{2+\sqrt[3]{2+\sqrt[4]{2+\cdots}}}<\sqrt{2+1.5213797}=1.876534$$
For a better upper bound we find:
$$\sqrt[4]{2+\sqrt[4]{2+\cdots}}=\frac{1}{3} \left(1-2\sqrt[3]{\frac{2}{47+3 \sqrt{249}}}+\sqrt[3]{\frac{47+3 \sqrt{249}}{2}} \right)=1.3532099$$
$$\sqrt{2+\sqrt[3]{2+\sqrt[4]{2+\cdots}}}<\sqrt{2+\sqrt[3]{2+1.3532099}}=1.8699639$$
So, we get, at this stage:

$$1.8685<\sqrt{2+\sqrt[3]{2+\sqrt[4]{2+\cdots}}}<1.8700$$

We get the value with uncertainty less than $0.1 \%$ in just a couple of steps.
We won't get a closed form this way, but this is much more accurate, than just truncating the nested radical.
</s>