Answer
(a) $0,0,0,0,0$
(b) $b_n=0$ for all $n$, so the sequence converges to $0$
(c) No, it oscillates between $\pm 1$ and $0$ .
Work Step by Step
(a) To evaluate $b_1, b_2, b_3, b_4$, and $b_5$, we substitute the given values of $n$ into the expression for $b_n=f(2 n+1)$ :
$\begin{align*} b_1 &= f(2(1)+1) = f(3) = \cos(\frac{3\pi}{2}) = 0 \\\ b_2 &= f(2(2)+1) = f(5) = \cos(\frac{5\pi}{2}) = 0 \ \\b_3 &= f(2(3)+1) = f(7) = \cos(\frac{7\pi}{2}) = 0 \ \\b_4 &= f(2(4)+1) = f(9) = \cos(\frac{9\pi}{2}) = 0 \ \\b_5 &= f(2(5)+1) = f(11) = \cos(\frac{11\pi}{2}) = 0 \ \end{align*}$
(b) To determine if $\left\{b_n\right\}$ converges, we need to check if it is a Cauchy sequence. Let's consider the difference between two arbitrary terms of the sequence:
$\begin{align*}
|b_n - b_m| &= |f(2n+1) - f(2m+1)| \\\
&= |\cos(\frac{\pi}{2}(2n+1)) - \cos(\frac{\pi}{2}(2m+1))| \\\
&= |\cos((n+\frac{1}{2})\pi) - \cos((m+\frac{1}{2})\pi)| \\\
&= 2|\sin(\frac{(n-m)\pi}{2})|\sin(\frac{(n+m+1)\pi}{2})|
\end{align*}$
Since $\sin x$ is bounded by $1$, we have:
Since the difference between any two terms is bounded above by a constant, we can conclude that $\left\{b_n\right\}$ is a Cauchy sequence. However, since the limit of $b_n$ is always 0 , the sequence converges to 0.
(c) To determine if $\{f(n)\}$ converges, we need to check if it is a Cauchy sequence. Let's consider the difference between two arbitrary terms of the sequence:
$\begin{align*}
|f(n) - f(m)| &= |\cos(\frac{\pi}{2}n) - \cos(\frac{\pi}{2}m)| \\\
&= 2|\sin(\frac{\pi(n+m)}{4})\sin(\frac{\pi(n-m)}{4})|
\end{align*}$
Since $\left|\sin \left(\frac{\pi(n+m)}{4}\right)\right|$ and $\left|\sin \left(\frac{\pi(n-m)}{4}\right)\right|$ are both bounded by $1$, we have:
Since the difference between any two terms is bounded above by a constant, we can conclude that $\{f(n)\}$ is a Cauchy sequence. However, since $f(n)$ oscillates between $- 1,0$ and $1$, depending on the parity of $n$, the sequence does not converge.
