Let \(\sum_{n=1}^{\infty}a_n\) be an absolutely convergent series. We'll verify Cauchy's criterion. Fix \(\varepsilon>0.
\exists N\geq 1\) so that if \(n\geq m\geq N\) we have \(\sum_{k-m}^n|a_k|<\varepsilon\text{.}\) Now for all \(\geq m\geq N,|\sum_{k=m}^n a_k|\leq \sum_{k=m}^n|a_k|<\varepsilon\text{.}\) So \(\sum_{n=1}^{\infty}\) satisfies Cauchy's criterion and hence converges.
Theorem3.6.
Suppose \(\sum_{n=1}^{\infty}a_n\) and \(\sum_{n=1}^{\infty}b_n\) are series and \(\sum_{n=1}^{\infty}b_n\) converges absolutely. If \(|a_n|\leq |b_n|\) for all \(n\geq 1\text{,}\) then \(\sum_{n=1}^{\infty}a_n\) converges absolutely.
Use Cauchy's criterion as above noting that if \(n\geq m\geq 1\text{,}\) then \(\sum_{k=m}^n |a_k|\leq\sum_{k=m}^n |b_k|\text{.}\)
Remark3.7.
The same result holds if one replaces \(|a_n|\leq |b_n|\) for all \(n\geq 1\) with \(\exists C\geq 0, N\geq 1\) such that \(|a_n|\leq C|b_n|\,\forall n\geq N\text{.}\)
Corollary3.8.
Suppose \(\sum_{n=1}^{\infty}a_n\) and \(\sum_{n=1}^{\infty}b_n\text{,}\) and \(\sum_{n=1}^{\infty}b_n\) converges absolutely and \(b_n\neq 0\) for all \(n\geq 1\text{.}\) (really it would be enough for the bns to nonzero after a point but...) If \(\limsup_{n\rightarrow\infty}|\frac{a_n}{b_n}|<\infty\text{,}\) then \(\sum_{n=1}^{\infty}a_n\) converges absolutely.
It's enough to produce \(C>0\) and \(N\geq 1\) so that \(|a_n|<C|b_n|\) for all \(n\geq N\) (then can use comparison test). Let \(C\in \mathbb{R}, C>\limsup_{n\rightarrow\infty}|\frac{a_n}{b_n}|\text{.}\) Then \(\exists N\geq 1\) so that for all \(n\geq N, |\frac{a_n}{b_n}|\leq C\text{.}\) So \(|a_n|\leq C|b_n|\) for all \(n\geq N\text{.}\)
Theorem3.9.
Suppose \(a,r\in\mathbb{C}\) and \(a\neq 0\text{.}\) Then \(\sum_{n=0}^{\infty}ar^n\) converges absolutely if \(|r|<1\) and diverges if \(|r|\geq 1\text{.}\) When \(|r|<1, \sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}\text{.}\)
If \(|r|\geq 1\) then \(|ar^n|\geq |a|\text{,}\) so \(\lim_{n\rightarrow\infty} ar^n\neq 0\) so \(\sum_{n=1}^{\infty}ar^n\) diverges. Suppose \(|r|<1\text{.}\) Set \(S_n=\sum_{k=0}^{n}ar^k\text{.}\) Then \(a+rS_n=S_{n+1}=S_n+ar^{n+1}\implies S_n=\frac{a(1-r^{n+1})}{1-r}\text{.}\) Now \(|S_n-\frac{a}{1-r}|=|\frac{a(a-r^{n+1})-a}{1-r}|=|\frac{ar^{n+1}}{a-r}|=\frac{|a|}{|a-r|}|r|^{n+1}\rightarrow_{n\rightarrow\infty}0\) (since \(|r|<1\)) so \(\sum_{n=0}^{\infty}ar^n=\lim_{n\rightarrow\infty} S_n=\frac{a}{1-r}\text{.}\)
Fix \(r\in \mathbb{R}\) with \(\limsup_{n\rightarrow\infty}|a_n|^{\frac{1}{n}}<r<1\text{.}\) Exists \(N\geq 1\) such that \(\forall n\geq N,|a_n|^{\frac{1}{n}}<r\) so \(|a_n|\leq r^n\forall n\geq N\text{.}\) Since \(\sum_{n=0}^{\infty}r^n\) converges (as \(r<1\)) we have \(\sum_{r=0}^{\infty}a_n\) converges absolutely by comparison test.
There is a subsequence \((a_{n_k})_{k=1}^{\infty}\) of \((a_{n})_{n=1}^{\infty}\) such that \(\lim_{k\rightarrow\infty}|a_{n_k}|^{\frac{1}{n_k}}=\limsup_{n\rightarrow\infty}|a_n|^{\frac{1}{n}}>1\text{.}\) Then exists \(K\geq 1\) such that for all \(k\geq K\text{,}\)\(|a_{n_k}|^{\frac{1}{n_k}}>1\text{,}\) and so \(|a_{n_k}|>1\) for all \(k\geq K\text{.}\) In particular, \(\limsup_{n\rightarrow\infty}|a_n|>1\text{,}\) so \((a_n)_{n=1}^{\infty}\) doesn't converge to 0. So \(\sum_{n=1}^{\infty}a_n\) diverges.
Stated without proof because proof is basically the same as the one above. Except here's a sketch: Fix \(r\in\mathbb{R}\text{,}\)\(\limsup_{n\rightarrow\infty}\frac{|a_{n+1}|}{|a_n|}<r<1\) and then exists \(N\geq 1\) such that for all \(n\geq N,\frac{|a_{n+1}|}{|a_n|}<r\text{.}\) Use this to show \(|a_n|\leq |a_N|r^{n-N}\forall n\geq N\text{.}\)
Theorem3.12.
If \((a_n)_{n=1}^{\infty}\subseteq \mathbb{R}_{\geq 0}\) is decreasing with \(\lim_{n\rightarrow\infty}a_n=0\) and \((b_n)_{n=1}^{\infty}\subseteq\mathbb{C}\) with \(\sup_{n\geq1}|\sum_{k=1}^{n}b_k|<\infty\text{,}\) then \(\sum_{n=1}^{\infty}a_nb_n\) converges.
Suppose \(a_n\geq 0\text{,}\)\(\lim_na_n=0, a_{n+1}\leq a_n\) for all \(n\geq 1\) and \((b_n)_{n=1}^{\infty}\subseteq \mathbb{C}\text{,}\)\(C:=\sup_{n\geq 1}|\sum_{k=1}^nb_k|<\infty\text{.}\) Set \(S_n=\sum_{k=1}^na_kb_k\) and \(r_n=\sum_{k=1}^nb_k\text{.}\) Claim: \(S_n=a_nr_n+\sum_{k=1}^{n-1}r_k(a_k-a_{k+1})\)
Now it's enough to show that (i) \((a_nr_n)_{n=1}^{\infty}\) converges and (ii) \(\sum_{k=1}^{\infty}r_k(a_k-a_{k-1})\) converges. For (i), \(|a_nr_n|\leq C|a_n|\rightarrow 0\text{,}\) so \(\lim_{n\rightarrow\infty}a_nr_n=0\text{.}\) Note that \(\left|{r_k(a_k-a_{k+1})}\right|\leq C(a_k-a_{k+1})\text{.}\) So for \(N\geq 1,\sum_{k=1}^N\left|{r_k(a_k-a_{k+1})}\right|\leq C\sum_{k=1}^N(a_k-a_{k+1})\) this is a telescoping sum so \(=C(a_1-a_{N+1})\rightarrow_{N\rightarrow \infty} Ca_1\text{.}\) Then \(\sum_{k=1}^N\left|{r_k(a_k-a_{k+1})}\right|\) is bounded above so it has to converge and hence \(\sum_{k=1}^N r_k(a_k-a_{k+1})\) converges, proving (ii).
Corollary3.13.Leibniz's Alternating Series Test.
If \((a_n)_{n=1}^{\infty}\subseteq \mathbb{R}_{\geq 0}\) is decreasing with \(\lim_{n\rightarrow\infty}a_n=0\text{,}\) then \(\sum_{n=1}^{\infty}(-1)^na_n\) converges.
