261 lines
6.6 KiB
Markdown
261 lines
6.6 KiB
Markdown
# Lecture 21
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## Review
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Recall the alternating series test from calculus: "Suppose $(a_n)^\infty_{n=1}$ is a sequence satisfies the following conditions:
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1. The sequence is nonnegative. (For all $n\in \mathbb{N}$, $a_n\geq 0$.)
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2. The sequence is decreasing. ($a_1\geq a_2\geq a_3\geq \cdots$)
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3. $\lim_{n\to\infty}a_n=0$.
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Then $\sum_{n=1}^\infty (-1)^{n+1}a_n$ converges."
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Exercise: Show that the statement above is false if we remove the second condition.
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[Hint: Use the fact that $\sum_{n=1}^\infty \frac{1}{n}$ diverges.]
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Let the sequence $a_n$ be defined as $a_n=\frac{1}{n},a_{n+1}=0$ for all $n\in \mathbb{N}$. This sequence satisfies the 1,3 but not the 2.
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And the harmonic series is not convergent.
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## New Material
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### Other tests for convergence of series
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Recall the integration by parts formula:
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Let $A(t),a(t),b(t)$ be functions of $t$ and $A'(t)=a(t)$.
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Then
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$$
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\begin{aligned}
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\int_p^q a(t)b(t)\,dt&=\int_p^q b(t)A'(t)\,dt\\
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&=\left.b(t)A(t)\right|_p^q-\int_p^q A(t)b'(t)\,dt
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\end{aligned}
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$$
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#### Theorem 3.41 Summation by parts
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Let $a_n,b_n$ be sequences.
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Let $A(n)=\sum_{k=1}^n a_k$. ($A_{-1}=0$). If $0\leq p\leq q$, then
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$$
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\sum_{n=p}^q a_nb_n=A_q b_q-A_{p-1}b_p-\sum_{n=p}^{q-1}A_n (b_n-b_{n+1})
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$$
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Proof:
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$$
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\begin{aligned}
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\sum_{n=p}^q a_nb_n&=\sum_{n=p}^q (A_n-A_{n-1})b_n\\
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&=\sum_{n=p}^q A_nb_n-\sum_{n=p}^q A_{n-1}b_n\\
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&=\sum_{n=p}^q A_nb_n-\sum_{n=p-1}^{q-1}A_n b_{n+1}\\
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&=A_qb_q-A_{p-1}b_p-\sum_{n=p}^{q-1} A_nb_n-\sum_{n=p}^{q-1} A_n b_{n+1}\\
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&=A_qb_q-A_{p-1}b_p-\sum_{n=p}^{q-1} A_n (b_n-b_{n+1})
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\end{aligned}
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$$
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QED
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#### Theorem 3.42 (Dirichlet's test)
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Suppose
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(a) the partial sum $A_n$ of $\sum a_n$ form a bounded sequence.
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(b) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing)
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(c) $\lim_{n\to\infty}b_n=0$.
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Then $\sum a_nb_n$ converges.
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Proof:
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By Cauchy criterion, it's enough to prove
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$\forall \epsilon >0, \exists N\in \mathbb{N}$ such that for all $p\geq q\geq N$,
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$$
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\left|\sum_{n=p}^q a_nb_n\right|<\epsilon
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$$
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By the partial sum $A_n$ of $\sum a_n$ form a bounded sequence. Let $\left|A_n\right|\leq M$ for all $n\in \mathbb{N}$.
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$$
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\begin{aligned}
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\left|\sum_{n=p}^q a_nb_n\right|&=\left|A_qb_q-A_{p-1}b_p-\sum_{n=p}^{q-1}A_n (b_n-b_{n+1})\right|\\
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&\leq |A_qb_q|+|A_{p-1}b_p|+\sum_{n=p}^{q-1}|A_n (b_n-b_{n+1})|\\
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&\leq M|b_q|+M|b_p|+\sum_{n=p}^{q-1}M(b_n-b_{n+1})\\
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&=M|b_q|+M|b_p|+M\sum_{n=p}^{q-1}(b_n-b_{n+1})\\
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&=M|b_q|+M|b_p|+M(b_p-b_q)\\
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&=2M|b_p|
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\end{aligned}
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$$
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Then we let $\epsilon >0$ be given. Since $b_n\to 0$, there exists $N\in \mathbb{N}$ such that for all $n\geq N$, $|b_n|<\frac{\epsilon}{2M}$.
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If $q\geq p\geq N$, then
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$$
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\left|\sum_{n=p}^q a_nb_n\right|\leq 2M|b_p|<\epsilon
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$$
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So $\sum a_nb_n$ converges.
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QED
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#### Theorem 3.43 (Alternating series test)
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Let $(b_n)^\infty_{n=1}$ be a sequence such that:
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(a) $b_1\geq b_2\geq b_3\geq \cdots$ (non-increasing)
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(b) $\lim_{n\to\infty}b_n=0$
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Then $\sum_{n=1}^\infty (-1)^{n+1}b_n$ converges.
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Proof:
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Let $a_n=(-1)^{n+1}$
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$A_n=\sum_{k=1}^n a_k=1$ if $n$ is odd, $0$ if $n$ is even.
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So $|A_n|\leq 1$ for all $n\in \mathbb{N}$.
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By Theorem 3.42, $\sum_{n=1}^\infty a_n b_n$ converges.
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QED
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Example:
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Consider the power series $\sum_{n=0}^\infty \frac{z^n}{n}$.
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The radius of convergence is $1$.
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We claim that the series converges for all $z\in \mathbb{C}$ with $|z|=1$ and $z\neq 1$.
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#### Theorem 3.44 Abel's test
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Let $(b_n)^\infty_{n=0}$ be a sequence such that:
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(a) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing)
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(b) $\lim_{n\to\infty}b_n=0$
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Then if $|z|=1$ and $z\neq 1$, $\sum_{n=0}^\infty b_nz^n$ converges.
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Proof:
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Fix $z\in \mathbb{C}$ with $|z|=1$ and $z\neq 1$.
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Let $a_n=z^n$.
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Then $A_n=\sum_{k=0}^n z^k=\frac{1-z^{n+1}}{1-z}$._
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$|A_n|\leq \frac{|1-z^{n+1}|}{|1-z|}$ for all $n\in \mathbb{N}$.
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By triangle inequality, $|1-z^{n+1}|\leq |1|+|z^{n+1}|=1+|z^{n+1}|$.
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And since $|z|=1$, $|z^{n+1}|=|z|^{n+1}=1$.
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So $|1-z^{n+1}|\leq 2$.
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So $|A_n|\leq \frac{2}{|1-z|}$ for all $n\in \mathbb{N}$.
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By Dirichlet's test, $\sum_{n=0}^\infty b_nz^n$
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QED
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### Absolute convergence
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The series $\sum_{n=0}^\infty a_n$ is said to **converge absolutely** if $\sum_{n=0}^\infty |a_n|$ converges.
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If $\sum_{n=0}^\infty a_n$ converges but does not converge absolutely, then $\sum_{n=0}^\infty a_n$ is said to **converge conditionally**.
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_Absolute convergence are nice but conditionally convergent series are not._
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#### Theorem 3.45 (Absolute convergence)
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If $\sum_{n=0}^\infty a_n$ converges absolutely, then $\sum_{n=0}^\infty a_n$ converges.
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Proof:
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Use comparison test.
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$$
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\sum_{n=0}^\infty |a_n|\geq \sum_{n=0}^\infty a_n
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$$
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QED
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Rearrangement of series:
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Let $f:\mathbb{N}\to \mathbb{N}$ be a bijection.
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If $\sum_{n=0}^\infty a_n$ is a sequence and $b_n=a_{f(n)}$, then $(b_n)^\infty_{n=0}$ is a rearrangement of $(a_n)^\infty_{n=0}$.
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If $\sum_{n=0}^\infty a_n$ converges absolutely, then any rearrangement of $\sum_{n=0}^\infty a_n$ converges to the same sum.
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Example:
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$a_n=\frac{(-1)^{n+1}}{n}$. $b_n=a_{f(n)}$.
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|n|1|2|3|4|5|6|7|8|9|
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|:-:|:-:|:-:|:-:|:-:|:-:|:-:|:-:|:-:|:-:|
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|$f(n)$|1|2|4|3|6|8|5|10|12|
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|$b_n$|1|-1/2|-1/4|1/3|-1/6|-1/8|1/5|-1/10|-1/12|
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$$
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\sum_{n=1}^\infty a_n=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\log 2
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$$
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$$
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\begin{aligned}
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\sum_{n=1}^\infty b_n&=1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\frac{1}{10}-\frac{1}{12}+\cdots\\
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&=\left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right)-\frac{1}{8}+\left(\frac{1}{5}-\frac{1}{10}\right)-\frac{1}{12}+\cdots\\
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&=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots\\
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&=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\right)\\
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&=\frac{1}{2}\log 2
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\end{aligned}
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$$
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You cannot always rearrange series.
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But, if $\sum_{n=0}^\infty a_n$ converges absolutely, then you can rearrange the series.
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#### Theorem 3.55
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Let $(a_n)^\infty_{n=0}$ be a sequence in $\mathbb{C}$ such that $\sum_{n=0}^\infty |a_n|$ converges absolutely.
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Then any rearrangement of $\sum_{n=0}^\infty a_n$ converges absolutely to the same sum.
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$$
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\sum_{n=0}^\infty a_n=\sum_{n=0}^\infty a_{f(n)}
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$$
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Ideas of proof:
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Let $f:\mathbb{N}\to \mathbb{N}$ be a bijection.
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and let $b_n=a_{f(n)}$.
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Let $s_n=\sum_{k=0}^n a_k,t_n=\sum_{k=0}^n b_k=\sum_{k=0}^n a_{f(k)}$.
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$I_n=\{1,2,\cdots,n\}$.
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$J_n=\{f(1),f(2),\cdots,f(n)\}$.
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$$
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\begin{aligned}
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s_n-t_n&=\sum_{k=0}^n a_k-\sum_{k=0}^n a_{f(k)}\\
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&=\sum_{k\in I_n} a_k-\sum_{k\in J_n} a_k\\
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&= \sum_{k\in I_n\setminus J_n} a_k+\sum_{k\in J_n\setminus I_n} a_k\\
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&\leq \sum_{k\in I_n\setminus J_n} |a_k|+\sum_{k\in J_n\setminus I_n} |a_k|
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\end{aligned}
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$$
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Key observation:
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For every $n\in \mathbb{N}$, there exists a $p$ such that $\{1,2,\cdots,n\}\subset I_n\cap J_n$.
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Then $|s_n-t_n|\leq \sum_{k=N+1}^\infty |a_k|$.
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QED
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