18. Sequences

\(\displaystyle \lim_{n\to\infty}a_n=L\) means For all \(\varepsilon \gt 0\), there is a positive integer, \(N\), such that if \(n \gt N\) then \(|a_n-L| \lt \varepsilon\). \(\displaystyle \lim_{n\to\infty}a_n=\infty\) means For all \(M \gt 0\), there is a positive integer, \(N\), such that if \(n \gt N\) then \(a_n \gt M\).

e2. Precise Special Limits

We now use the precise definition of a limit to prove the various theorems involving limits. We start with the four special limits.

\[ (1) \qquad \lim_{n\to\infty}n=\infty \]

Identity Sequence Since \(a_n=n\) and we want to show the limit is infinite, we need to show:

For all \(M \gt 0\), there is a positive integer, \(N\), such that

if \(n \gt N\) then \(n \gt M\).

So given an arbitrary number \(M \gt 0\), we let \(N\) be any integer bigger than \(M\), i.e. \(N \gt M\). Then, trivally,

if \(n \gt N\) then \(n \gt N \gt M\).

For any number \(c\), \[ (2) \qquad \lim_{n\to\infty}c=c \]

Constant Sequences Since \(a_n=c\) and we want to show the limit is the finite number \(L=c\), we need to show:

For all \(\varepsilon \gt 0\), there is a positive integer, \(N\), such that

if \(n \gt N\) then \(|c-c| \lt \varepsilon\).

So given an arbitrary number \(\varepsilon \gt 0\), we let \(N=1\). Then, trivally,

if \(n \gt 1\) then \(|c-c|=0 \lt \varepsilon\).

If \(p>0\), then \[ (3a) \qquad \lim_{n\to\infty}n^p=\infty \qquad \text{and} \qquad (3b) \qquad \lim_{n\to\infty}\dfrac{1}{n^p}=0. \] If \(p<0\), then \[ (3c) \qquad \lim_{n\to\infty}n^p=0 \qquad \text{and} \qquad (3d) \qquad \lim_{n\to\infty}\dfrac{1}{n^p}=\infty. \]
(3c) is the same as (3b). (3d) is the same as (3a).

Power Sequences (3a) Since \(a_n=n^p\) with \(p \gt 0\) and we want to show the limit is infinite, we need to show:

For all \(M \gt 0\), there is a positive integer, \(N\), such that

if \(n \gt N\) then \(n^p \gt M\).

So given an arbitrary number \(M \gt 0\), we let \(N\) be any integer bigger than \(M^{1/p}\), i.e. \(N \gt M^{1/p}\). Then,

if \(n \gt N\) then \(n^p \gt N^p \gt \left(M^{1/p}\right)^p=M\).

How did we know to take \(N \gt M^{1/p}\)? Answer: We work backwards from what we want: \[ n^p \gt M \quad \Longleftarrow \quad n \gt M^{1/p} \] which will follow if \(n \gt N\) and \(N \gt M^{1/p}\).

Power Sequences (3b) Since \(a_n=\dfrac{1}{n^p}\) with \(p \gt 0\) and we want to show the limit is the finite number \(L=0\), we need to show:

For all \(\varepsilon \gt 0\), there is a positive integer, \(N\), such that

if \(n \gt N\) then \(\left|\dfrac{1}{n^p}\right| \lt \varepsilon\).

So given an arbitrary number \(\varepsilon \gt 0\), we let \(N\) be any integer bigger than \(\varepsilon^{-1/p}\), i.e. \(N \gt \varepsilon^{-1/p}\). Then,

if \(n \gt N\) then \(|n^p| \gt |N^p| \gt \left(\varepsilon^{-1/p}\right)^p=\varepsilon^{-1}\) and so \(\left|\dfrac{1}{n^p}\right| \lt \varepsilon\).

How did we know to take \(N \gt \varepsilon^{-1/p}\)? Answer: We work backwards from what we want: \[ \left|\dfrac{1}{n^p}\right| \lt \varepsilon \quad \Longleftarrow \quad |n^p| \gt \varepsilon^{-1} \quad \Longleftarrow \quad |n| \gt \varepsilon^{-1/p} \] which will follow if \(n \gt N\) and \(N \gt \varepsilon^{-1/p}\).

If \(b>1\), then \[ (4a) \qquad \lim_{n\to\infty}b^n=\infty \qquad \text{and} \qquad (4b) \qquad \lim_{n\to\infty}\dfrac{1}{b^n}=0. \] If \(0 < b < 1\), then \[ (4c) \qquad \lim_{n\to\infty}b^n=0 \qquad \text{and} \qquad (4d) \qquad \lim_{n\to\infty}\dfrac{1}{b^n}=\infty. \]
(4c) is the same as (4b). (4d) is the same as (4a).

Exponential Sequences (4a) Since \(a_n=b^n\) with \(b \gt 1\) and we want to show the limit is infinite, we need to show:

For all \(M \gt 0\), there is a positive integer, \(N\), such that

if \(n \gt N\) then \(b^n \gt M\).

So given an arbitrary number \(M \gt 0\), we let \(N\) be any integer bigger than \(\log_b(M)\), i.e. \(N \gt \log_b(M)\). Then,

if \(n \gt N\) then \(b^n \gt b^N \gt b^{\log_b(M)}=M\).

How did we know to take \(N \gt \log_b(M)\)? Answer: We work backwards from what we want: \[ b^n \gt M \quad \Longleftarrow \quad n \gt \log_b(M) \] which will follow if \(n \gt N\) and \(N \gt \log_b(M)\).

Exponential Sequences (4b) Since \(a_n=\dfrac{1}{b^n}\) with \(b \gt 1\) and we want to show the limit is the finite number \(L=0\), we need to show:

For all \(\varepsilon \gt 0\), there is a positive integer, \(N\), such that

if \(n \gt N\) then \(\left|\dfrac{1}{b^n}\right| \lt \varepsilon\).

So given an arbitrary number \(\varepsilon \gt 0\), we let \(N\) be any integer bigger than \(\log_b(\varepsilon^{-1})\), i.e. \(N \gt \log_b(\varepsilon^{-1})\). Then,

if \(n \gt N\) then \(|b^n| \gt |b^N| \gt b^{\log_b(\varepsilon^{-1})}=\varepsilon^{-1}\) and so \(\left|\dfrac{1}{b^n}\right| \lt \varepsilon\).

How did we know to take \(N \gt \log_b(\varepsilon^{-1})\)? Answer: We work backwards from what we want: \[ \left|\dfrac{1}{b^n}\right| \lt \varepsilon \quad \Longleftarrow \quad |b^n| \gt \varepsilon^{-1} \quad \Longleftarrow \quad n \gt \log_b(\varepsilon^{-1}) \] which will follow if \(n \gt N\) and \(N \gt \log_b(\varepsilon^{-1})\).

18. Sequences

e2. Precise Special Limits

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