# Sequences and Series/Print version

The current, editable version of this book is available in Wikibooks, the open-content textbooks collection, at
https://en.wikibooks.org/wiki/Sequences_and_Series

# Multiple limits

Theorem (interchanging summation and integration):
Let $(\Omega ,{\mathcal {F}},\mu )$ be a measure space, and let $(f_{n})_{n\in \mathbb {N} }$ be a sequence of functions from $\Omega$ to $\mathbb {K} ^{d}$, where $\mathbb {K} =\mathbb {R}$ or $\mathbb {C}$. If either of the two expressions
    $\int _{\Omega }\sum _{n=1}^{\infty }\|f_{n}(\omega )\|_{\infty }\mu (d\omega )$ or $\sum _{n=1}^{\infty }\int _{\Omega }\|f_{n}(\omega )\|_{\infty }\mu (d\omega )$
converges, so does the other, and we have
    $\int _{\Omega }\sum _{n=1}^{\infty }f_{n}(\omega )\mu (d\omega )=\sum _{n=1}^{\infty }\int _{\Omega }f_{n}(\omega )\mu (d\omega )$.
Proof: Regarding the summation as integration over $\mathbb {N}$ with σ-algebra $2^{\mathbb {N} }$ and counting measure, this theorem is an immediate consequence of Fubini's theorem, given that integration and summation are defined pointwise. $\Box$
Theorem (interchanging summation and real differentiation):
Let $(f_{n})_{n\in \mathbb {N} }$ be a sequence of continuously differentiable functions from an open subset $U$ of $\mathbb {R} ^{d}$ to $\mathbb {R} ^{k}$. Suppose that both
    $\sum _{n=1}^{\infty }\|f_{n}(x)\|_{\infty }$ and $\sum _{n=1}^{\infty }\|Df_{n}(x)\|_{\infty }$
converge for all $x\in U$, and that for all $x\in U$ there exists $\delta >0$ and a sequence $(a_{n})_{n\in \mathbb {N} }$ in $\mathbb {R}$ such that
    $\sum _{n=1}^{\infty }a_{n}<\infty$ and $\forall n\in \mathbb {N} :\forall y\in {\overline {B_{\delta }(x)}}:a_{n}\geq |Df_{n}(y)|$.
Then
    $D\sum _{n=1}^{\infty }f_{n}(x)=\sum _{n=1}^{\infty }Df_{n}(x)$
for all $x\in U$.
Proof: $\Box$

# Series and integration

Theorem (Abelian partial summation):
Let $(a_{n})_{n\in \mathbb {N} }$ be a sequence of complex numbers, and let $f:[1,\infty )\to \mathbb {C}$ be differentiable on $(1,\infty )$. Finally define
    $A(x):=\sum _{1\leq n\leq x}a_{n}$.
Then for $x\geq 1$ we have
    $\sum _{1\leq n\leq x}a_{n}f(n)=A(x)f(x)-\int _{1}^{x}A(t)f'(t)dt$.
Proof: If $m=\lfloor x\rfloor$, we have
    ${\begin{aligned}\int _{1}^{x}A(t)f'(t)dt&=\sum _{k=2}^{m}\int _{k-1}^{k}A(t)f'(t)dt+\int _{m}^{x}A(t)f'(t)dt\\&=\sum _{k=2}^{m}\sum _{1\leq n\leq k-1}a_{n}\int _{k-1}^{k}f'(t)dt+\sum _{1\leq n\leq m}a_{n}\int _{m}^{x}f'(t)dt\\&=\sum _{k=2}^{m}A(k-1)(f(k)-f(k-1))+A(x)f(x)-A(m)f(m).\end{aligned}}$
But
    ${\begin{aligned}\sum _{k=2}^{m}A(k-1)(f(k)-f(k-1))&=\sum _{k=2}^{m}A(k-1)f(k)-\sum _{k=2}^{m}A(k-1)f(k-1)\\&=\sum _{k=2}^{m}A(k-1)f(k)-\sum _{k=1}^{m-1}A(k)f(k)\\&=\sum _{k=2}^{m-1}f(k)(\underbrace {A(k-1)-A(k)} _{=-a_{k}})+A(m-1)f(m)-A(1)f(1).\end{aligned}}$
so that
    $\int _{1}^{x}A(t)f'(t)dt=A(x)f(x)-\sum _{k=1}^{m}a_{k}f(k)$. $\Box$

# Power series

Proposition (identity theorem for one-dimensional power series):
Let
    $f(z):=\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}$ and $g(z):=\sum _{n=0}^{\infty }b_{n}(z-z_{0})^{n}$
be two (complex or real) power series that converge on $B_{\epsilon }(z_{0})$ for some $\epsilon >0$. Suppose that $z_{0}$ is an accumulation point of the set $\{z\in B_{\epsilon }(z_{0})|f(z)=g(z)\}$. Then we have $a_{n}=b_{n}$ for all $n\in \mathbb {N} _{0}$.
Proof: Assume that not $a_{n}=b_{n}$ for all $n\in \mathbb {N} _{0}$. Then there exists a least $n$ (call it $n_{0}$) such that $a_{n_{0}}\neq b_{n_{0}}$. Consider the function
    $h(z):=f(z)-g(z)=\sum _{n=0}^{\infty }(a_{n}-b_{n})(z-z_{0})^{n}$,
which is defined on at least $B_{\epsilon }(z_{0})$. Since $a_{n}=b_{n}$ for $n<n_{0}$, the power series $h$ starts at $(z-z_{0})^{n_{0}}$. Therefore,
    $j(z):={\frac {h(z)}{(z-z_{0})^{n_{0}}}}=\sum _{n=n_{0}}(a_{n}-b_{n})(z-z_{0})^{n-n_{0}}$
is a well-defined function on $B_{\epsilon }(z_{0})$ which is also continuous due to the continuity of power series. Moreover,
    $j(0)=a_{n}-b_{n}\neq _{0}$,
and by continuity of $|j(z)|$, there exists a $\delta >0$ such that $|j(z)|>0$ for all $z\in B_{\delta }(z_{0})$. But by definition,
    $h(z)=(z-z_{0})^{n}j(z)$,
so that we have for $z\in B_{\delta }(z_{0})\setminus \{z_{0}\}$ that $|h(z)|=|z-z_{0}|^{n}|j(z)|>0$ and consequently $h(z)\neq 0$, and hence $f(z)\neq g(z)$. But this contradicts the assumption that $z_{0}$ was an accumulation point of $\{z\in B_{\epsilon }(z_{0})|f(z)=g(z)\}$. $\Box$
Example (falsity of the identity theorem for multi-dimensional power series):
For multi-dimensional power series, that is power series of the type
    $h(z):=\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }(z-z_{0})^{\alpha }$ for a $z_{0}=(z_{0,1},\ldots ,z_{0,d})\in \mathbb {C} ^{d}$,
the set $\{z|h(z)=0\}$ may have $z_{0}$ as an accumulation point even when $h$ does not vanish. An easy example (which works in any dimension $d\geq 2$) is $z_{0}=0$ and
    $h(z)=z_{1}z_{2}$.
Theorem (Abel's theorem):
Let
    $\sum _{n=1}^{\infty }a_{n}z^{n}$
be a real or complex power series of convergence radius $1$, and suppose that
    $\lim _{z\to 1}\sum _{n=1}^{\infty }a_{n}z^{n}=\alpha \in \mathbb {C}$.
Then
    $\sum _{n=1}^{\infty }a_{n}=\alpha$.
{{proof|By Abelian partial summation, we have
    $\sum _{1\leq n\leq x}a_{n}z^{n}=z^{x}A(x)-\ln(z)\int _{1}^{x}A(t)z^{t}dt$
for $|z|<1$ and $x\geq 1$, where we denote as usual
    $A(x):=\sum _{1\leq n\leq x}a_{n}$.
Substituting $z=\exp(w)$, we get
    $\sum _{1\leq n\leq x}a_{n}\exp(wn)=\exp(wx)A(x)-w\int _{1}^{x}A(t)\exp(wt)dt$.
We then put $$

# Dirichlet‒Hurwitz series

Definition (Dirichlet‒Hurwitz series):
Let $f:\mathbb {N} \to \mathbb {C}$ be a function, and let $a\in \mathbb {C} \setminus \{-1,-2,\ldots \}$. The Dirichlet‒Hurwitz series associated to $f$ and $a$ is the function of $s\in \mathbb {C}$ given by the series
    $\sum _{n=1}^{\infty }{\frac {f(n)}{(n+a)^{s}}}$.
Definition (abscissa of absolute convergence of Dirichlet‒Hurwitz series):
Let $f:\mathbb {N} \to \mathbb {C}$ be a function, and let $a\in \mathbb {C} \setminus \{-1,-2,\ldots \}$. Suppose that there exists a number $\sigma _{a}\in \mathbb {R}$ such that
    $\sum _{n=1}^{\infty }\left|{\frac {f(n)}{(n+a)^{s}}}\right|$
converges whenever $\Re s>\sigma _{a}$ and diverges whenever $\Re s<\sigma _{a}$. Then $\sigma _{a}$ is called the abscissa of absolute convergence of the Dirichlet‒Hurwitz series associated to $f$ and $a$.
Proposition (existence of abscissa of absolute convergence of Dirichlet‒Hurwitz series):
Let $f:\mathbb {N} \to \mathbb {C}$ be a function, and let $a\in \mathbb {C} \setminus \{-1,-2,\ldots \}$. Suppose that

# Infinite products

Definition (infinite product):
Let $(b_{n})_{n\in \mathbb {N} }$ be a sequence of numbers in $\mathbb {K} =\mathbb {R}$ or $\mathbb {C}$. If the limit
    $\lim _{N\to \infty }\prod _{n=1}^{N}b_{n}$
exists, it is called the infinite product of $(b_{n})_{n\in \mathbb {N} }$ and denoted by
    $\prod _{n=1}^{\infty }b_{n}$.
Proposition (necessary condition for convergence of infinite products):
In order for the infinite product
    $\prod _{n=1}^{\infty }b_{n}$
of a sequence $(b_{n})_{n\in \mathbb {N} }$ to exist and not to be zero, it is necessary that
    $\lim _{n\to \infty }b_{n}=1$.
Proof: Suppose that not $\lim _{n\to \infty }b_{n}=1$. Then there exists $\epsilon >0$ and an infinite sequence $(n_{k})_{k\in \mathbb {N} }$ such that for all $k\in \mathbb {N}$ we have $|b_{n}-1|>\epsilon$. Thus, upon denoting
    $P_{N}:=\prod _{n=1}^{N}b_{n}$,
we will have
    $|P_{n_{k}}-P_{n_{k}-1}|=|P_{n_{k}-1}||b_{n}|$.
Suppose for a contradiction that $\lim _{N\to \infty }P_{N}$ exited and was equal to $c\in \mathbb {R}$. Then when $k$ is sufficiently large, we will have
    $\left||P_{n_{k}-1}||b_{n}|-|c|\right|\geq |c|\epsilon /2$,
which is a contradiction. $\Box$
Proposition (series criterion for the convergence of infinite products):
Let $(a_{n})_{n\in \mathbb {N} }$ be a sequence of real numbers. If
    $\sum _{n=1}^{\infty }|a_{n}|<\infty$,
then
    $\prod _{n=1}^{\infty }(1+a_{n})$
converges.
Proof: $\Box$