Step 1: Understand the sequence.
Each term is a nested square root of $7$. The first term is $a_1=\sqrt{7}$, the next is $a_2=\sqrt{7+\sqrt{7}}$, and so on. Every new term adds one more layer. All terms are positive.
Step 2: Find what value it heads toward.
If the nesting goes on forever, the value settles at a number $L$ that satisfies $L=\sqrt{7+L}$. This is because adding one more layer to a very deep nest does not change it much.
Step 3: Solve for that limit.
Square both sides: $L^2=7+L$, so $L^2-L-7=0$. Using the quadratic formula, $L=\frac{1+\sqrt{1+28}}{2}=\frac{1+\sqrt{29}}{2}$. We keep the positive root since the value is positive.
Step 4: Estimate the limit.
Since $\sqrt{29}\approx 5.385$, the limit is about $\frac{1+5.385}{2}\approx 3.19$. So the whole sequence climbs toward roughly $3.19$.
Step 5: Show terms stay below the limit.
Check the first term: $a_1=\sqrt{7}\approx 2.65$, which is below $3.19$. If any term $a_n$ is below $L$, then $a_{n+1}=\sqrt{7+a_n}<\sqrt{7+L}=L$. So by this chain reaction every term stays under $L$.
Step 6: Show the sequence keeps rising.
Each term is bigger than the one before, but never reaches $L$. So the terms increase and are squeezed under the ceiling $\frac{1+\sqrt{29}}{2}$.
Step 7: State the true bound.
Therefore for every $n$ the term is strictly less than this limit value.
\[ \boxed{a_n<\dfrac{1+\sqrt{29}}{2}\ \ \forall n\ge 1} \]