Even more limits of sequences
Define the sequence \((b_{n})\) by
\[b_{n} = \left\{\begin{array}{ll} 0 & \text{if \(n\) is even}\\ \frac{1}{n^2} & \text{if \(n\) is odd}\end{array}\right.\]
Prove that \(\displaystyle{\lim (b_{n})=0}\).
Scratch work. We want to figure out how large \(n\) must be so that
\(|b_{n}|<\varepsilon\), but this depends on whether \(n\) is even or odd. So, we consider two cases:
If \(n\) is even, then \(b_{n}=0\) and hence \(|b_{n}|=0<\varepsilon\). So \(n\) could be any positive integer.
If \(n\) is odd, then \(|b_{n}| = \frac{1}{n^2}\leq \frac{1}{n}\). So it is enough that \(n>\frac{1}{\varepsilon}\).
Thus, it is enough to take \(K(\varepsilon)>\frac{1}{\varepsilon}\).
Proof. Let \(\varepsilon>0\) be given. By the Archimedean Property there is a natural number \(K(\varepsilon)> \frac{1}{\varepsilon}\). Let \(n\in\N\) such that \(n\geq K(\varepsilon)\) be given. Note that \(n> \frac{1}{\varepsilon}\), and hence \(\frac{1}{n}<\varepsilon\). We will consider two cases:
If \(n\) is even, then
\[|b_{n} - 0| = 0<\varepsilon.\]
If \(n\) is odd, then
\[|b_{n} - 0| = \frac{1}{n^2}\leq \frac{1}{n}<\varepsilon.\]
In either case, we have shown that for an arbitrary natural number
\(n\geq K(\varepsilon)\) we have \(|b_{n}-0|<\varepsilon.\) \(\Box\)
Define the sequence \((b_{n})\) by
\[b_{n} = \left\{\begin{array}{ll} 0 & \text{if \(n\) is even}\\ \frac{1}{n^2} & \text{if \(n\) is odd}\end{array}\right.\]
Prove that \(\displaystyle{\lim (b_{n})=0}\).
Prove that \(\displaystyle{\lim \left(\frac{1}{n}\right)\neq 1}\).
This is the negation of the statement \(\displaystyle{\lim \left(\frac{1}{n}\right)= 1}\), that is, for all \(\varepsilon>0\) there exists \(K\in\N\) such that for all\(n\geq K\),
\[\left|\frac{1}{n} - 1\right|<\varepsilon.\]
The negation can be stated as follows:
There exists \(\varepsilon>0\) such that for all \(K\in\N\), there exists \(n\geq K\) such that
\[\left|\frac{1}{n} - 1\right|\geq \varepsilon.\]
Prove that \(\displaystyle{\lim \left(\frac{1}{n}\right)\neq 1}\).
Proof. Set \(\varepsilon=\frac{1}{2}\). Let \(K\in\N\) be arbitrary. Let \(n\) be a natural number greater than \(K\) which is also greater than \(3\). Since \(n>3\), we see that \(\frac{1}{n}<\frac{1}{3}<\frac{1}{2}\). This implies \(1-\frac{1}{n}>\frac{1}{2}=\varepsilon\). Therefore
\[\left|\frac{1}{n} - 1\right| = \left|\frac{1-n}{n}\right| = \frac{n-1}{n} = 1-\frac{1}{n}>\varepsilon.\ \Box\]
There exists \(\varepsilon>0\) such that for all \(K\in\N\), there exists \(n\geq K\) such that
\[\left|\frac{1}{n} - 1\right|\geq \varepsilon.\]
Proof. Set \(\varepsilon=1\). Let \(K\in\N\) be arbitrary. Let \(n\) be an odd natural number greater than \(K\). Since \(n\) is odd and \(n\geq 1\) we see that
\[(-1)^n+\frac{1}{n} = -1+\frac{1}{n}\leq -1+1=0.\]
Adding \(-1\) to both sides we have
\[(-1)^n+\frac{1}{n}-1\leq -1.\]
Using this we deduce
\[\left|(-1)^n+\frac{1}{n} - 1\right|\geq 1.\ \Box\]
Prove that \(\displaystyle{\lim\left((-1)^n+\frac{1}{n}\right)\neq 1}\).
Practice problem.
State the precise meaning of the statement \(x\) is not the limit of the sequence \((x_{n})\).
State the precise meaning of the statement \((x_{n})\) does not have a limit.
There exists \(\varepsilon>0\) such that for all \(K\in\N\), there exists \(n\geq K\) such that \(|x_{n}-x|\geq\varepsilon\).
For all \(x\in\R\) there exists \(\varepsilon>0\) such that for all \(K\in\N\) there exists \(n\geq K\) such that \(|x_{n}-x|\geq \varepsilon\).