Question

Help with sequences

Answer 1

For (a), you're essentially proving by induction that \(a_n\) is bounded above by 2.

Base case: \(n=1\).
\[a_1=\sqrt2\le2\]
(true because it's given.)
Assume this is true for \(n=k\), i.e.
\[a_k\le2\]
Now show that means it's also true for \(n=k+1\), i.e.
\[a_{k+1}\le2\]
From the definition of the sequence and the assumption,
\[a_{k+1}=\sqrt{2+a_k}\le\sqrt{2+\sqrt2}\le2\]
Thus proving boundedness.

Answer 2

To prove monotonicity, you must show that \(a_n\le a_{n+1}\) for all \(n\). Again, by induction, your base case is
\[a_1=\sqrt 2\\
a_2=\sqrt{2+\sqrt 2}\]
Obviously, \(a_1<a_2\), proving the base case.

Assume this is true for \(n=k\) (so that \(a_{k+1}>a_k\)), and show it holds for \(n=k+1\):
\[a_{k+1}=\sqrt{2+a_{k}}\\
a_{k+2}=\sqrt{2+a_{k+1}}>\sqrt{2+a_k}=a_{k+1}\]
thus proving monotonicity.

Answer 3

Now you know that \(a_n\) converges, since it's bounded and monotonic, meaning the limit exists. Call it \(L\). Now,
\[\begin{align*}\lim_{n\to\infty}a_n&=\lim_{n\to\infty}a_{n+1}=L\\
L&=\lim_{n\to\infty}\sqrt{2+a_n}\\
L&=\sqrt{2+L}\\
L^2&=2+L\\
L^2-L-2&=0\\
L&=2,-1
\end{align*}\]
Ignore the negative root; the limit must be positive because this sequence is bounded above by a positive number and is increasing toward it.