Series

Series are the sum of Sequences, where we take a sequence $(a_n{)}_{n=1}^{\infty }$ and add up all of its terms.

Convergence §

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Series”,“label”:“series”,“_index”:0}] Definition 1 (Series). A series $(a_n{)}_{n=1}^{\infty }$ converges if the sequence of partial sums $S_1=a_1,S_2=a_1+a_2,S_3=a_1+a_2+a_3\dots$ converges. We then set ${\sum }_{k=1}^{\infty }{a}_k={\lim }_{n\to \infty }{S}_n$.

Examples §

Show that

$$\sum _{k=0}^{\infty }{x}^k=\frac{1}{1-x},|x|<1$$

\begin{proof}

Let

$$S_n=\sum _{k=0}^n{x}^k=x^0+x^1+\cdots +x^n$$

Then,

$$x{S}_n=x\sum _{k=0}^n{x}^k=x^1+x^2+\cdots +x^{n+1}.$$

Subtracting the second equation from the first, we get

$$\begin{array}{c}{S}_n-S_{n}x=1-x^{n+1}\\ S_n(1-x)=1-x^{n+1}\\ S_n=\frac{1-x^{n+1}}{1-x}.\end{array}$$

Now, by the definition of convergence, we just need to make sure that ${\lim }_{n\to \infty }{S}_n$ converges. Plugging in, we get

$$\underset{n\to \infty }{\lim }\frac{1-x^{n+1}}{1-x}=\frac{1}{1-x}.$$

Therefore,

$$x{S}_n=x\sum _{k=0}^n{x}^k=x^1+x^2+\cdots +x^{n+1}$$

converges. \end{proof}

Sum of Two Converging Series §

[!math|{“type”:“lemma”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Sum of 2 Converging Series”,“label”:“sum-of—converging-series”,“_index”:1}] Lemma 2 (Sum of 2 Converging Series). If $\sum a_k$ and $\sum b_k$ converge, then $\sum (\alpha a_k+\beta b_k)$ converges for $\alpha ,\beta \in \mathbb{R}$.

Divergence of Harmonic Series §

[!math|{“type”:“lemma”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Harmonic Series”,“label”:“harmonic-series”,“_index”:2}] Lemma 3 (Harmonic Series). ${\sum }_{k=1}^{\infty }\frac{1}{k}$ diverges.

\begin{proof}

Observe that

$$\begin{array}{rl}\sum _{k=n+1}^{2n}\frac{1}{k}& =\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{2n}\\ & >\frac{1}{2n}+\frac{1}{2n}+\cdots +\frac{1}{2n}\\ & =\frac{n}{2n}=\frac{1}{2}.\end{array}$$

Therefore,

$$\begin{array}{rl}\sum _{k=1}^{2^S}\frac{1}{k}& =\sum _{k=1}^2\frac{1}{k}+\sum _{k=3}^4\frac{1}{k}+\sum _{k=5}^8\frac{1}{k}+\cdots +\sum _{k=2^{S-1}+1}^{2^S}\frac{1}{k}\\ & >\frac{S}{2}\end{array}$$

Thus, the set of partial sums diverges. \end{proof}

Telescoping Series §

Similarly, ${\sum }_{k=1}^{\infty }\frac{1}{k+1}$ diverges. However, their difference ${\sum }_{k=1}^{\infty }\frac{1}{k}-\frac{1}{k+1}$ converges. This is because of telescoping series.

$$\begin{array}{rl}\sum _{k=1}^{\infty }\frac{1}{k}-\frac{1}{k+1}& =\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\cdots -\frac{1}{\infty }\\ & =1\end{array}$$

Necessary Limit Condition §

[!math|{“type”:“lemma”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Limit of Converging Series Term”,“label”:“limit-of-converging-series-term”,“_index”:3}] Lemma 4 (Limit of Converging Series Term). If ${\sum }_{k=1}^{\infty }{a}_n$ converges, then ${\lim }_{n\to \infty }{a}_n=0$.

\begin{proof}

$$S_n=\underset{n\to \infty }{\lim }\sum _{k=1}^n{a}_k=L⟹S_{n-1}=L⟹S_n-S_{n-1}=L-L=0=a_n$$

\end{proof}

Comparison Test §

If ${\sum }_{}^{}{b}_n$ converges and $a_n\le c{b}_n$ for $n\ge 1,c>0$, then ${\sum }_{}^{}{a}_n$ converges.

\begin{proof}This is obvious when comparing partial sums (both are bounded). \end{proof}

Limit Comparison Test §

If $a_n,b_n>0$ and ${\lim }_{n\to \infty }\frac{a_n}{b_n}=1$, then ${\sum }_{}^{}{a}_n$ converges iff ${\sum }_{}^{}{b}_n$ converges.

\begin{proof}${\lim }_{n\to \infty }\frac{a_n}{b_n}=1⟹\exists N\in \mathbb{N}$ such that $\frac{1}{2}<\frac{a_n}{b_n}<\frac{3}{2}$ for $n\ge N⟹b_n<2a_n,a_n<\frac{3}{2}{b}_n$ for $n\ge N$. Then, we can apply Comparison Test. \end{proof}

Root Test §

If $a_k\ge 0$ for $k\ge 1$ in ${\sum }_{k=1}^{\infty }{a}_k$ and ${\lim }_{n\to \infty }{a}_k^{\frac{1}{k}}=R$, then

1. If $R<1$, the series converges ($a_n=\frac{n}{3^n}$)
2. If $R>1$, the series diverges ($a_n=n^2{2}^n$)
3. If $R=1$, inconclusive ($a_n=\frac{1}{n\log n}$)

\begin{proof} If $R<1$, choose $x$ such that $R<x<1$. Then, $\exists N\mid a_n^{\frac{1}{n}}<x$ for $n\ge N$. Then, $a_n<x^n$ for $n\ge N$ and $x<1⟹x^n\to 0$.

If $R>1$, then ${\lim }_{n\to \infty }{a}_n\ne 0$. \end{proof}

Ratio Test §

Suppose

$$\sum _{k=1}^{\infty }{a}_k:\forall k\ge 1,a_k>0,\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=L.$$

1. If the limit $L<1$, the series converges
2. If $L>1$, the series diverges
3. If $L=1$, inconclusive

\begin{proof} If $L<1$, let $x:L<x<1$. Then, there exists $N$ such that $\frac{a_{n+1}}{a_n}<x$ for all $n\le N$. This implies $\frac{a_{n+1}}{x^{n+1}}<\frac{a_n}{x_n}$ for all $n\ge N$. Thus, $\frac{a_n}{x^n}<c$ for some $c>0$, and ${\sum }_{}^{}{a}_n$ converges by the Comparison Test.

If $L>1$, then $a_{n+1}>a_n$ for all $n\ge N$, which clearly doesn’t converge. \end{proof}

Absolute Convergence §

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Absolute Convergence”,“label”:“absolute-convergence”,“_index”:4}] Definition 5 (Absolute Convergence). ${\sum }_{}^{}{a}_n$ is absolutely convergent if ${\sum }_{}^{}|a_n|$ converges. If ${\sum }_{}^{}{a}_n$ converges and ${\sum }_{}^{}|a_n|$ diverges, then ${\sum }_{}^{}{a}_n$ is conditionally convergent.

Absolute convergence is useful because

Absolute Convergence Test §

[!math|{“type”:“lemma”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:5}] Lemma 6. If ${\sum }_{}^{}{a}_n$ is absolutely convergent, it is convergent.

\begin{proof} Let $b_n=a_n+|a_n|$. Then $bn=0$ or $bn=2|a_n|$, so $0\le b_n\le 2|a_n|$. By Comparison Test, ${\sum }_{}^{}{b}_n$ converges, so ${\sum }_{}^{}{a}_n={\sum }_{}^{}{b}_n-{\sum }_{}^{}|a_n|$ also converges. \end{proof}

To find conditional convergence, we use

Alternating Series Test §

If $a_n$ is monotonically decreasing and $a_n\to 0$, then ${\sum }_{}^{}(-1{)}^{n-1}{a}_n$ converges.

\begin{proof} The partial sums $S_{2n}$ are monotonic increasing since

S_{2n+2} - S_{2n} = a_{2n+1} - a_{2n+2} > 0 .$$ Similarly, $S_{2n+1}$ is decreasing. Both sequences are bounded below by $S_{2}$ and above by $S_{1}$. Then,

\begin{align} \lim_{ n \to \infty } S_{2n} - \lim_{ n \to \infty } S_{2n-1} &= \lim_{ n \to \infty } S_{2n}-S_{2n-1} \ &= \lim_{ n \to \infty } -a_{2n} \ &= 0 \end{align}

Thus, $\sum_{}^{}a_{n}$ converges. `\end{proof}` ## Integral Test For $S=\sum_{n=1}^{\infty}a_{n}$ w/ $a_{n}=f(n)$ for $f>0$, monotone and decreasing on $[1,\infty]$, then $S$ converges iff $I=\int_{1}^{\infty} f(x) \, dx$ converges. `\begin{proof}`Partition $[1,N]$ into $1<2<3<\dots<N$ so

\begin{align} U(f,P_{N})&=a_{1}+a_{2}+\dots+a_{N-1}\ L(f,P_{N})&=a_{2} + \dots + a_{N} .\end{align}

$$Weknow$$

L(f,P_{N})<\int_{1}^{N} f , dx < \int_{1}^{\infty} f(x) , d\mathbf{x}.

Let $N\to \infty$, so $S-a_{1}<\int_{1}^{\infty} f(x) \, dx$. For the converse, notice

\int_{1}^{N} f(x) , dx =\sum_{j=1}^{j=N-1}\int_{j}^{j+1} f(x) , dx.

Let the $b_{j}=\int_{j}^{j+1} f(x) \, dx$. Clearly, $a_{n+1}<b_{n}\leq a_{n}$, so by [[#comparison-test|Comparison Test]], $S-a_{1}\leq\int_{1}^{\infty} f(x) \, dx\leq S$. `\end{proof}` # Power Series > [!math|{"type":"definition","number":"auto","setAsNoteMathLink":false,"title":"Power Series","label":"power-series","_index":6}] Definition 7 (Power Series). > A power series is an infinite series of the form > $$ > \sum_{n=0}^{\infty}a_{n}(z-a)^{n}=a_{0}+a_{1}(z-a)+a_{2}(z-a)^{2}+\dots > $$ ## Convergence > [!math|{"type":"lemma","number":"auto","setAsNoteMathLink":false,"_index":7}] Lemma 8. > If the power series $\sum_{}^{}a_{n}z^{n}$ converges for a particular $z=z_{1}\neq 0$, then it converges absolutely for all $z:|z|<|z_{1}|$. ^d5c99e `\begin{proof}` Since $\sum_{}^{}a_{n}z_{1}^{n}$ converges, $a_{n}z_{1}^{n}\to0$ as $n\to \infty$. Therefore, there is $N \in \mathbb{N}$ s.t. $|a_{n}z_{1}^{n}|<1$ for all $n\geq N$. If $|z|<|z_{1}|$, then

|a_{n}z^{n}|=|a_{n}z_{1}^{n}| |\frac{z}{z_{1}}|^{n}< |\frac{z}{z_{1}}|^{n}

for all $n\geq N$. Since $\sum_{}^{}|\frac{z}{z_{1}}|^{n}<\infty$, $\sum_{}^{}a_{n}z^{n}$ converges by comparison. `\end{proof}` ## Radius of Convergence > [!math|{"type":"axiom","number":"auto","setAsNoteMathLink":false,"title":"Radius of Convergence","label":"radius-of-convergence","_index":8}] Axiom 9 (Radius of Convergence). > If the $\sum_{}^{}a_{n}z^{n}$ converges for at least one $z=z_{1}\neq 0$ and diverges for $z=z_{2}\neq 0$, there exists a positive real number $R$ called the radius of convergence s.t. the series converges absolutely if $|z|<R$ and diverges if $|z|>R$. `\begin{proof}` Let $A$ be the set of $|z|$ for which $\sum_{}^{}a_{n}z^{n}$ converges. Then $A\neq \emptyset$ and the set is bounded by $|z_{2}|$. Therefore, $R=\text{sup} A$ exists. Then, $\sum_{}^{}a_{n}z^{n}$ diverges for all $|z|>R$. If $|z|<R$, then $\exists x$ with $|z|<x<R$ s.t. the series converges. So by [[#^d5c99e|^d5c99e]], $a_{n}z^{n}$ also converges. `\end{proof}` We find the radius of convergence by using the [[Series#ratio-test|Ratio Test]].