Proving $\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}} < 3$ for $n\geq 1$ by induction How can I prove by induction that $\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}} < 3$ for $n\geq 1$? My guess is that there must be another form to express the sum of nested square roots, but I don't know how to find it.

We can actually prove a more general version of what you hope to prove (set $a=0$ for your problem, specifically):
Claim: For every $n\in\mathbb{Z^+}$ and every non-negative real number $a$
$$
\sqrt{a+1+\sqrt{a+2+\cdots+\sqrt{a+n}}}< a+3.
$$
Proof. For $n\geq 1$, let $S(n)$ denote the statement that for any non-negative real $a$,
$$
S(n) : \sqrt{a+1+\sqrt{a+2+\cdots+\sqrt{a+n}}}< a+3.
$$
Base step: $S(1)$ says that $\sqrt{a+1}<a+3$, and this is verifiable since
$$
a+1<(a+3)^2\Leftrightarrow 0<a^2+5a+8,
$$
which is true for $a\geq 0$. 
Inductive step: Fix some $k\geq 0$, and suppose that $S(k)$ is true for any non-negative $a$ where
$$
S(k) : \sqrt{a+1+\sqrt{a+2+\cdots+\sqrt{a+k}}}< a+3.
$$
To be shown is that $S(k+1)$ follows for any non-negative $b$ where
$$
S(k+1) : \underbrace{\sqrt{b+1+\sqrt{b+2+\cdots+\sqrt{b+k+\sqrt{b+k+1}}}}}_{\text{LHS}}<\underbrace{b+3}_{\text{RHS}}.
$$
Using $a=b+1$,
\begin{align}
\text{LHS} &= \sqrt{b+1+\sqrt{b+2+\cdots+\sqrt{b+k+\sqrt{b+k+1}}}}\\[1em]
&= \sqrt{b+1+\sqrt{a+1+\cdots+\sqrt{a+k-1+\sqrt{a+k}}}}\\[1em]
&< \sqrt{b+1+a+3}\tag{by $S(k)$}\\[0.5em]
&= \sqrt{2b+5}\\[0.5em]
&< b+3\\[0.5em]
&= \text{RHS},
\end{align}
where the last inequality follows since
$$
2b+5<(b+3)^2=b^2+6b+9
$$
and $b^2\geq 0$. This proves $S(k+1)$, completing the inductive step.
By mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$

Your particular problem holds for $S(n)$ where $a=0$. 
