4. Series

Def.: ${\sum }_{k=1}^{\infty }{a}_k$ for some sequence $a_n$

Sequence of partial sums: $s_n:={\sum }_{k=0}^n{a}_k$

Absolutely convergent: $\sum |a_n|$ converges Conditionally convergent: $\sum a_n$ converges but $\sum |a_n|$ does not

Necessary Condition for Convergence

Lemma

If a series converges, then $\underset{n\mapsto \infty }{\lim }{a}_n=0$

Not sufficient: $a_n=\frac{1}{n}$ converges, but its corresponding series does not. Intuitively: terms shrink, but not fast enough to limit the sum. Reason: We can group the terms:

$$1+\underset{\ge \frac{1}{2}}{\underbrace{\frac{1}{2}}}+\underset{\ge \frac{1}{2}}{\underbrace{\frac{1}{3}+\frac{1}{4}}}+\underset{\ge \frac{1}{2}}{\underbrace{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}}}+\underset{\ge \frac{1}{2}}{\underbrace{\frac{1}{9}+\cdots +\frac{1}{16}}}+\cdots$$

Riemann Rearrangement Theorem

Theorem

If $\sum a_n$ is conditionally convergent and $L\in \mathbb{R}$ is any real number, then there exists a bijection $\phi :{\mathbb{N}}_0\to {\mathbb{N}}_0$ s.t.

$$\sum _{n=0}^{\infty }{a}_{\phi (n)}=L$$

Proof:

• For it to be conditionally convergent, $a_n$ must contain infinitely many positive and negative terms (an inf. large pool).
• We first add positive terms until we overshoot $L$. We then add negative terms until we undershoot $L$. We repeat this process into infinity.
• We over/undershoot by at most the size of the last term. But since $a_n\to 0$, both $p_n\to 0$ and $q_n\to 0$. Hence, we overshoot and undershoot ever closer to $L$, eventually converging.

Absolutely convergent series: This does not apply! We don’t have the infinite pools. By the Absolute Series Rearrangement Theorem, ${\sum }_{n=0}^{\infty }{a}_n={\sum }_{n=0}^{\infty }{a}_{\phi (n)}$

Cauchy Criterion

Partial sums are sequences $⟹$ can be [[3. Sequences#3. Sequences#Cauchy Sequence|Cauchy Sequence]] $⟹$ converge iff Cauchy Sequence

We can simplify the def since the terms $k=1\dots m$ cancel:

$$\forall \epsilon >0,\exists N\text{ s.t. }m,n\ge N⟹|S_n-S_m|=\left|\sum _{k=m+1}^n{a}_k\right|\le \epsilon$$

Cauchy Product

Theorem

Let ${\sum }_{n=0}^{\infty }{a}_n$ and ${\sum }_{n=0}^{\infty }{b}_n$ be absolutely converging series. Then:

$$\left(\sum _{n=0}^{\infty }{a}_n\right)\left(\sum _{n=0}^{\infty }{b}_n\right)=\sum _{n=0}^{\infty }\left(\sum _{k=0}^n{a}_{n-k}{b}_k\right)$$

The series to the right converges absolutely (not series to the left).

Absolute Convergence Implies Convergence

Theorem

Every absolutely convergent series converges, and the generalized triangle inequality holds:

$\left|{\sum }_{n=0}^{\infty }{a}_n\right|\le {\sum }_{n=0}^{\infty }|a_n|$

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Leibnitz Test

Theorem

Let $(a_n{)}_{n\in {\mathbb{N}}_0}$ be a monotonically decreasing sequence of non-negative real numbers converging to zero.

Then the alternating series ${\sum }_{n=0}^{\infty }(-1{)}^n{a}_n$ converges, and:

$$\sum _{n=0}^{2m+1}(-1{)}^n{a}_n\le \sum _{n=0}^{\infty }(-1{)}^n{a}_n\le \sum _{n=0}^{2m}(-1{)}^n{a}_n\forall m\in {\mathbb{N}}_0$$

Proof:

• Even: $S_{2m+2}=S_{2m}-\underset{\ge 0}{\underbrace{(a_{2m+1}-a_{2m+2})}}\le S_{2m}$ ($a_n$ mono decreasing)
• Odd: $S_{2m+3}=S_{2m+1}+\underset{\ge 0}{\underbrace{(a_{2m+2}-a_{2m+3})}}\ge S_{2m+1}$

Comparison Test (Vergleichskriterium)

Theorem

Let ${\sum }_{n=0}^{\infty }{a}_n$, ${\sum }_{n=0}^{\infty }{b}_n$, and ${\sum }_{n=0}^{\infty }{c}_n$ have non-negative terms. Suppose $\exists N\in \mathbb{N}$ s.t.

$$c_n\le b_n\le a_n\forall n\ge N$$

• ${\sum }_{n=0}^{\infty }{a}_n$ converges $⟹{\sum }_{n=0}^{\infty }{b}_n$ converges (Majorant Criterion)
• ${\sum }_{n=0}^{\infty }{c}_n$ diverges $⟹{\sum }_{n=0}^{\infty }{b}_n$ diverges (Minorant Criterion)

Quotient Test (Quotienten-Kriterium)

Theorem

$(a_n{)}_{n\in {\mathbb{N}}_0}\in \mathbb{R}$. Then:

$$\rho =\underset{n\to \infty }{\lim }\frac{|a_{n+1}|}{|a_n|}$$

• $\rho <1$ $⟹{\sum }_{n=0}^{\infty }{a}_n$ converges absolutely
• $\rho >1$ $⟹{\sum }_{n=0}^{\infty }{a}_n$ does not converge
• $\rho =1⟹$ the test is inconclusive

Root Test (Wurzelkriterium)

Same as Quotient Test, but with $\rho =\underset{n\to \infty }{\mathrm{limsup}}\left(|a_n{|}^{1/n}\right)\in \mathbb{R}\cup \{\infty \}$

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Power Series

Def: $c_0+c_1(x-a)+c_2(x-a{)}^2+c_3(x-a{)}^3+\dots$ $=\sum _{k=0}^{\infty }{c}_k(x-a{)}^k$

Radius of convergence:

• the series converges for $x$ with $|x-a|<R$ and
• the series diverges for $x$ with $|x-a|>R$.

Interval of convergence: $(a-R,a+R)$

Lemma

The radius of convergence $R$ can be computed with the following formula:

$$\rho =\underset{n\to \infty }{\mathrm{limsup}}|c_n{|}^{1/n},$$ $$R=\{\begin{array}{ll}0,& \text{if }\rho =\infty \\ {\rho }^{-1},& \text{if }0<\rho <\infty \\ \infty ,& \text{if }\rho =0\end{array}$$

Tips

• $\sum _{n=1}^{\infty }\frac{1}{n^2}=\frac{{\pi }^2}{6}$

Partial Fractions (Telescoping)

For $a\ne b$:

$$\frac{1}{(n+a)(n+b)}=\frac{1}{b-a}\left(\frac{1}{n+a}-\frac{1}{n+b}\right)$$

Method: $\frac{A}{n+a}+\frac{B}{n+b}$, multiply through by $(n+a)(n+b)$, solve by plugging in $n=-a$ and $n=-b$.

Why useful: The resulting series telescopes — most terms cancel in the partial sums, leaving only boundary terms. This lets you compute exact values.