10.31 Convergence
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Algebra of limits:
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Let \(\lim_{n \to \infty} a_n = a\) and \(\lim_{n \to \infty} b_n = b\). Then
If \(a_n \leq b_n\) for all \(n\), then \(a \leq b\).
Proof
\(\lim_{n\to\infty}a_{n}=a\implies\forall\,\epsilon_1>0,\,\exists\,N_1\in\mathbb{N}\text{ such that }|a_{n}-a|<\epsilon_1,\forall\,n\geq N_1\)
\(\lim_{n\to\infty}b_{n}=b\implies\forall\,\epsilon_2>0,\,\exists\,N_2\in\mathbb{N}\text{ such that }|b_{n}-b|<\epsilon_2,\forall\,n\geq N_2\)
Then \(-\epsilon_1<a_{n}-a<\epsilon_1\) and \(-\epsilon_2<b_{n}-b<\epsilon_2\)
Since \(a_n\leq b_n,\forall n\), then \(\left(a_{n}\right)_{\max}\leq\left(b_{n}\right)_{\min}\Rightarrow\varepsilon_1+a\leq-\varepsilon_2+b\)
Thus \(a-b\leq-\varepsilon_1-\varepsilon_2\)
We need to prove \(a\leq b\Rightarrow a-b\leq 0\)
Since \(-\varepsilon_1-\varepsilon_2<0\), then \(a-b\leq 0\)
Thus \(a \leq b\) 2. If \(\lim_{n \to \infty} a_n = 0\), and \(|b_{n}|<c\) for all \(n\), then \(\lim_{n \to \infty} a_n b_n = 0\).
Proof
Take \(\frac{\varepsilon}{c}>0\)
\(\lim_{n\to\infty}a_{n}=0\implies\exists\,N\in\mathbb{N}\text{ such that }|a_{n}-0|<\frac{\varepsilon}{c}\,,\forall\,n\geq N\)
\(|b_n| < c\)
\(|a_{n}b_{n}-0|=|a_{n}b_{n}|=|a_{n}||b_{n}|<\frac{\varepsilon}{c}\cdot c=\varepsilon\), take the same \(N\)
Then \(\lim_{n \to \infty} a_n b_n = 0\)
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We say that \(\lim_{n \to \infty} a_n = \infty\) if for any positive real number \(M\) there is a natural number \(N\) such that for every natural number \(n \geq N\), we have \(a_n > M\).
We say that \(\lim_{n \to \infty} a_n = -\infty\) if for any negative real number \(K\) there is a natural
number \(N\) such that for every natural number \(n \geq N\), we have \(a_n < K\).Prove that:
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If \(\lim_{n \to \infty} a_n = \infty\) and \(\lim_{n \to \infty} b_n = b \neq 0\) then \(\lim_{n \to \infty} a_n b_n =\begin{cases}\infty, & \text{if } b > 0 \\-\infty, & \text{if } b < 0\end{cases}\)
Proof
If \(b > 0\).
Since \(\lim_{n \to \infty} b_n = b\) and \(b > 0\), there exist \(N_1 \in \mathbb{N}\) such that \(b_{n}>\frac{b}{2}\) for all \(n \geq N_1\).Lemma 1: If \lim_{n \to \infty} a_n = \ell and \ell > 0, then there is a number X such th...
On the other hand, for \(M > 0\) , since \(\lim_{n \to \infty} a_n = \infty\), there exists \(N_2 \in \mathbb{N}\) such that for every natural number \(n \geq N_2\), we have \(a_n > M\).
The following statement is useful: if \(0 < x < y\) and \(0 < z < w\), then \(0 < xz < yw\).
Continuing with the proof, we now have:
\(b_{n}>\frac{b}{2}>0\) and \(a_n > M > 0\).
Then \(a_{n}b_{n}>\frac{Mb}{2}>0\) for \(N = \max \{N_1, N_2\}\). Thus let \(M^{\prime}=\frac{Mb}{2}\), we have \(a_nb_n>M'\).
Then \(\lim_{n \to \infty} a_n b_n = \infty\).
If \(b < 0\).
Since \(\lim_{n \to \infty} b_n = b\) and \(b < 0\), there exist \(N_1 \in \mathbb{N}\) such that \(b_{n}<\frac{b}{2}\) for all \(n \geq N_1\)
On the other hand, for \(M > 0\) , since \(\lim_{n \to \infty} a_n = \infty\), there exists \(N_2 \in \mathbb{N}\) such that for every natural number \(n \geq N_2\), we have \(a_n > M\).
Thus \(-b_{n}>-\frac{b}{2}>0\) and \(a_n > M > 0\).
Then \(-a_{n}b_{n}>-\frac{Mb}{2}>0\) for \(N = \max \{N_1, N_2\}\). Thus let \(M^{\prime}=\frac{Mb}{2}\), we have \(a_nb_n<M'\).
Then \(\lim_{n\to\infty}a_{n}b_{n}=-\infty\).
Professor's proof
We will use the following claim.
Claim 1: If \(\lim_{n \to \infty} b_n = b\) and \(b > 0\), then there exists \(\delta > 0\) and \(N \in \mathbb{N}\) such that \(b_n > \delta\) for all \(n \geq N\).
Proof of Claim 1:
\(\forall \, \epsilon > 0, \, \exists \, N \in \mathbb{N} \text{ such that } |b_n - b| < \epsilon \, \forall \, n \geq N\)
\(-\epsilon < b_n - b < \epsilon\)
\(-\epsilon + b < b_n < \epsilon + b\)
if \(0 < \epsilon < b \implies \delta = b - \epsilon > 0\), \(N \in \mathbb{N}\)
Then \(b_n > \delta\)
Since \(\lim_{n \to \infty} b_n = b\) and \(b > 0\), there exist \(0 < \delta\) and \(N_1 \in \mathbb{N}\) such that \(b_n > \delta\) for all \(n \geq N_1\).
On the other hand, for \(M > 0\) (*), since \(\lim_{n \to \infty} a_n = \infty\), there exists \(N_2 \in \mathbb{N}\) such that for every natural number \(n \geq N_2\), we have \(a_n > M\).
The following statement is useful: if \(0 < x < y\) and \(0 < z < w\), then \(0 < xz < yw\).
Continuing with the proof, we now have:
\(b_n > \delta > 0\) and \(a_n > M > 0\).
Then \(a_n b_n > \delta M > 0\) for \(N = \max \{N_1, N_2\}\). Now change \(M\) in (*) by \(M / \delta\), then \(\lim_{n \to \infty} a_n b_n = \infty\).
Now, consider the case \(b < 0\). We use the following claim:
Claim 2: If \(\lim_{n \to \infty} b_n = b\) and \(b < 0\), then there exists \(\delta < 0\) and \(N \in \mathbb{N}\) such that \(b_n < \delta\) for all \(n \geq N\).
Proof of Claim 2:
\(\forall \, \epsilon > 0, \, \exists \, N \in \mathbb{N} \text{ such that } |b_n - b| < \epsilon \, \forall \, n \geq N\)
\(-\epsilon < b_n - b < \epsilon\)
\(-\epsilon + b < b_n < \epsilon + b\)
if \(0>\epsilon>b\implies\delta=b-\epsilon<0\), \(N \in \mathbb{N}\)
Then \(b_{n}<\delta\)
Since \(\lim_{n \to \infty} b_n = b\) and \(b < 0\), there exist \(\delta < 0\) and \(N_1 \in \mathbb{N}\) such that \(b_n < \delta\) for all \(n \geq N_1\).
Suppose \(K < 0\) (**), then \(-K > 0\). Since \(\lim_{n \to \infty} a_n = \infty\), there exists \(N_2 \in \mathbb{N}\) such that for every natural number \(n \geq N_2\), we have \(a_n > -K\).
Thus, \(a_n > -K > 0\) and \(b_n < \delta < 0\) for \(N = \max \{N_1, N_2\}\). Therefore,
\(a_n > -K\)
\(a_n b_n < -K b_n\)
\(a_{n}b_{n}<-Kb_{n}-K\delta<0\)The last inequality holds for all \(n \geq N\). You can replace \(K\) in (**) with \(K / |\delta|\), then \(\lim_{n \to \infty} a_n b_n = -\infty\)
I think it is unnecessary
2. If \(\lim_{n \to \infty} a_n = -\infty\) and \(\lim_{n \to \infty} b_n = b \neq 0\) then \(\lim_{n \to \infty} a_n b_n =\begin{cases}-\infty, & \text{if } b > 0 \\\infty, & \text{if } b < 0\end{cases}\)
Similarly....
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Find the limits of the following sequences:
Trick
\(\lim_{n \to \infty} \frac{P(n)}{Q(n)} = ?\)
\(\frac{P(n)}{Q\left(n\right)}=\frac{a_{k}n^{k}+a_{k-1}n^{k-1}+\dots+a_1n+a_0}{b_{T}n^{T}+b_{T-1}n^{T-1}+\dots+b_1n+b_0}=\frac{n^{k}\left(a_{k}+\frac{a_{k-1}}{n}++\frac{a_1}{n^{k-1}}+\frac{a_0}{n^{k}}\right)}{n^{T}\left(b_{T}+\frac{b_{T-1}}{n}++\frac{b_1}{n^{T-1}}+\frac{b_0}{n^{T}}\right)}\)
\(\text{where } \deg(P) = k, \, a_k \neq 0\)
\(\text{and } \deg(Q) = T, \, b_T \neq 0\)
Case \(T > K\)
\(\lim_{n\to\infty}\frac{P(n)}{Q(n)}=\lim_{n\to\infty}\frac{n^{k}\left(a_{k}+\frac{a_{k-1}}{n}++\frac{a_1}{n^{k-1}}+\frac{a_0}{n^{k}}\right)}{n^{T}\left(b_{T}+\frac{b_{T-1}}{n}++\frac{b_1}{n^{T-1}}+\frac{b_0}{n^{T}}\right)}\)
\(=\lim_{n\to\infty}\frac{1}{n^{T-K}}\cdot\lim_{n\to\infty}\left(\frac{a_{k}+\frac{a_{k-1}}{n}+\dots+\frac{a_1}{n^{k-1}}+\frac{a_0}{n^{k}}}{b_{T}+\frac{b_{T-1}}{n}+\dots+\frac{b_1}{n^{T-1}}+\frac{b_0}{n^{T}}}\right)\)
Since \(\lim_{n \to \infty} \frac{1}{n^{T - K}} = 0\), it follows that \(\lim_{n \to \infty} \frac{P(n)}{Q(n)} = 0\)
Case \(T = K\)
\(\lim_{n\to\infty}\frac{P(n)}{Q(n)}=\lim_{n\to\infty}\frac{a_{k}+\frac{a_{k-1}}{n}+\dots+\frac{a_1}{n^{k-1}}+\frac{a_0}{n^{k}}}{b_{k}+\frac{b_{k-1}}{n}+\dots+\frac{b_1}{n^{k-1}}+\frac{b_0}{n^{k}}}=\frac{a_{k}}{b_{k}}\)
As \(n \to \infty\), all terms involving \(\frac{1}{n}\) in the numerator and denominator go to \(0\)
Case \(T < K\)
\(\lim_{n \to \infty} \frac{P(n)}{Q(n)} = \lim_{n \to \infty} n^{K - T} \left( \frac{a_k + \frac{a_{k-1}}{n} + \dots + \frac{a_1}{n^{k-1}} + \frac{a_0}{n^k}}{b_T + \frac{b_{T-1}}{n} + \dots + \frac{b_1}{n^{T-1}} + \frac{b_0}{n^T}} \right)\)
As \(n \to \infty\), \(n^{K - T} \to \infty\), so:
\(\lim_{n \to \infty} \frac{P(n)}{Q(n)} = \begin{cases} +\infty & \text{if } \frac{a_k}{b_T} > 0 \\ -\infty & \text{if } \frac{a_k}{b_T} < 0 \end{cases}\)
where \(a_k \neq 0\) and \(b_T \neq 0\)
- \((a_n) = \frac{n - 1}{n + 1}\) \(1\)
- \((b_n) = \frac{3n^2 + 2n + 2}{5n^2 + 4}\) \(\frac35\)
- \((c_n) = \frac{2n + 2}{5n^2 + 4}\) \(0\)
- \((d_n) = \frac{3n^2 + 2n + 2}{5n + 4}\) \(+\infty\)
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\((e_n) = \frac{2^n + n^2}{2^n + 5}\)
\(\lim_{n \to \infty} \frac{2^n + n^2}{2^n + 5} = \lim_{n \to \infty} \frac{2^n \left( 1 + \frac{n^2}{2^n} \right)}{2^n \left( 1 + \frac{5}{2^n} \right)}\)
Assume: \(\lim_{n \to \infty} \frac{P(n)}{a^n} = 0, \, a > 1\)
\(\lim_{n \to \infty} \frac{2^n + n^2}{2^n + 5} = \lim_{n \to \infty} \frac{1 + \frac{n^2}{2^n}}{1 + \frac{5}{2^n}} = 1\)
Or use Sandwich Theorem
\(1 \leq \frac{2^n + n^2}{2^n + 5} \leq 1 + \frac{1}{n}\) for \(n > 3\)
Since \(2^n + 5 \leq 2^n + n^2\), then \(1 \leq \frac{2^n + n^2}{2^n + 5}\)
Also \(\frac{2^n + n^2}{2^n + 5} \leq \frac{2^n + n^2}{2^n} = 1 + \frac{n^2}{2^n} \leq 1 + \frac{1}{n}\)
Claim: \(\frac{n^2}{2^n} \leq \frac{1}{n}\)
\(n^3 \leq 2^n\) for \(n \geq 10\)
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\(\lim_{n \to \infty} \sqrt[n]{a} = ?\), where \(a > 0\)
Claim: If \(1 < a\) and \(1 < b\), then \(a < b \iff a^n < b^n\)
\(\Leftarrow\)) N.T.P. \(a^n<b^n\Rightarrow a<b\)
\(a^{n}<b^{n}\begin{cases}a<b\rightarrow \text{ the only option}\\ a>b\rightarrow a^n>b^n\\ a=b\rightarrow a^n<b^n\end{cases}\)
To prove: \(a < b \implies a^n < b^n\) (Exercise: Use induction.)
Back to the proof
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Case \(a = 1\): \(\lim_{n \to \infty} \sqrt[n]{1} = 1\)
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Case \(a > 1\): \(\lim_{n \to \infty} \sqrt[n]{a} = 1\)
For \(1<a<a^2<\dots<a^{n}<a^{n+1}\Rightarrow1<a^{n\frac{n+1}{n+1}}<a^{n+1\frac{n}{n}}\Rightarrow1<(\sqrt[n+1]{a})^{n(n+1)}<(\sqrt[n]{a})^{n(n+1)}\)
Then \(1 < \sqrt[n]{a} < \sqrt[n+1]{a}\) as \(n \to \infty\)
Claim: \(\sqrt[n]{a}\) is a decreasing function for \(a > 1\) and is bounded below by \(1\).
Moreover, \(\forall n,1 < a^{\frac{1}{n}}\), then, by the theorem, the sequence converges to \(0<L\leq1\)
Let's \(a^{\frac{1}{n}} = a_n\) and consider a subsequence of
\(b_{n}=(a_{n})^2=\left(\sqrt[n]{a}\right)^2=\sqrt[n]{a}\cdot\sqrt[n]{a}\).
take n is even subsequence
\(\lim_{n\to\infty}b_{n}=\lim_{n\to\infty}\sqrt[n]{a}\cdot\lim_{n\to\infty}\sqrt[n]{a}=L\cdot L=L\)
\(L(1 - L) = 0 \implies L = 1 \text{ or } L = 0\).
Since \(L \leq 1\), we conclude that \(L = 1\).
- Case \(0 < a < 1\): \(\lim_{n \to \infty} \sqrt[n]{a} = 1\)
Since \(\frac{1}{a} > 1\), let \(c=\frac{1}{a}\).
Then \(c^{\frac{1}{n}}=\frac{1}{a^{\frac{1}{n}}}\).
\(\lim_{n\to\infty}c^{\frac{1}{n}}=\lim_{n\to\infty}\frac{1}{a^{\frac{1}{n}}}=1\).
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If \(a_n \geq 0\) and \(\lim_{n \to \infty} a_n = a \geq 0\), then \(\lim_{n \to \infty} \sqrt{a_n} = \sqrt{a}\).
- Case \(a = 0\): Use the definition.
We need to prove \(\lim_{n\to\infty}\sqrt{a_{n}}=0\), and we know \(\lim_{n\to\infty}a_{n}=0\).
Then \(\forall\varepsilon_1,\exists n_0,n>n_0,\)\(|a_{n}-0|<\varepsilon\Rightarrow\sqrt{a_{n}}<\sqrt{\varepsilon}\Rightarrow\left|\sqrt{a_{n}}-0\left|\right.<\sqrt{\varepsilon_1}\right.<\varepsilon_2\)
Then we can choose \(\varepsilon_2=\varepsilon_1^2\) * Case \(a > 0\):
\(N.T.P.\left|\sqrt{a_{n}}-\sqrt{a}\right|<\varepsilon\\\text{We want use}\quad|a_{n}-a|<\epsilon.\\\text{Then }|\sqrt{a_{n}}-\sqrt{a}|=\frac{|a_{n}-a|}{\left|\sqrt{a_{n}}+\sqrt{a}\right|}<\epsilon\times\frac{1}{\left|\sqrt{a_{n}}+\sqrt{a}\right|}<\epsilon\times\frac{1}{\left|\sqrt{a_{n}}\right|+\left|\sqrt{a}\right|}<\epsilon\times\frac{1}{M+\sqrt{a}}\\\text{Thus, we need to take }(M+\sqrt{a})\varepsilon\)
Example:
\(\lim_{n \to \infty} \sqrt{1 + \frac{1}{n}} = \sqrt{\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)} = \sqrt{1} = 1\).