\input /home/pasky/school/fastex/lib.tex


\subchapter{�ady s nez�porn�mi �leny}{

Studujeme \M{\sum a_{n}}, kde $a_{n} \ge 0$ pro $\forall n\in\nat$.


\scpart{Pozorov�n�}{
	Pro �ady s nez�porn�mi �leny mohou nastat pouze dv� mo�nosti.
	$\sum a_{n} \in \real$ nebo $\sum a_{n} = +\infty$.
	To plyne z toho, �e posloupnost $\{s_{m}\}_{n=1}^{\infty}$ je neklesaj�c�,
	a tedy dle \ias{2}{9}{} $\existss \lim s_{m} = \sum a_{n}$.
}

\theorem{2}{linearita konvergentn�ch �ad}{
	Nech� $\sum a_{n}$ a $\sum{b_{n}}$ konverguj�. Pak:
	\list{
		\listItem{\M{\sum(a_{n}+b_{n})} konverguje. }
		\listItem{\M{\sum{c_{n}} \textbox{konverguje} \Longleftrightarrow{} \sum{\a{}c_{n}} \textbox{konverguje}}
			pro $\forall \a \in{} \real{}\setminus{}\emptyset$.
		}
	}

	\shortdef{D�kaz}{\DCV}
}


\theorem{3}{srovn�vac� krit�rium}{

	Nech� $\sum{a_{n}}$ a $\sum{b_{n}}$ jsou �ady s nez�porn�mi �leny.
	Nech� d�le existuje $n_{0}\in{}\nat{}$ takov�, �e pro $\forall n\ge{}n_{0}$ plat� $a_{n}\le{}b_{n}$. Pak:

	\list{
		\listItem{\M{\sum{b_{n}} \textbox{konverguje} \Rightarrow{} \sum{a_{n}}} konverguje. }
		\listItem{\M{\sum{a_{n}} \textbox{diverguje} \Rightarrow{} \sum{b_{n}} } diverguje. }
	}

	Ob� tvrzen� ��kaj� vpodstat� tot�. ($a_{n}$ je ``hodn�j��'' �ada, $b_{n}$ ``zlobiv�j��''.)

	\proof{
	\list{
		\listItem{$s_{n} = a_{1} + \cdots + a_{n}$, $\s_{m} = b_{1} + \cdots + b_{m}$

			D�le definujme $ \s := \lim{\s_{m}} \in{} \real{}$.

			Nav�c v�me, �e $\{s_n\}$ je posloupnost neklesaj�c�.

			$$ s_{n} = a_{1} + \cdots + a_{n_{0}} + a_{n_{0}+1} + \cdots + a_{n} $$
			$$      \le{} a_{1} + \cdots + a_{n_{0}} + \s_{n}\qquad (\forall n\ge{}n_{0}) $$
			$$      \le{} a_{1} + \cdots + a_{n_{0}} + \s    \qquad (\in{} \real{}) $$

			Tedy je $\{s_{n}\}$ neklesaj�c� shora omezen� posloupnost, tud��
			$\sum{a_{n}}$ je dle \ias{2}{9}{} konvergentn�.
		}

		\listItem{$\Longleftrightarrow{}$ (i)}
	}

	\qed
	}

}


\penalty-10000

\theorem{4}{limitn� srovn�vac� krit�rium}{

	Nech� $\sum_{n=1}^{\infty}{a_{n}}$, $\sum_{n=1}^{\infty}{b_{n}}$ jsou �ady s nez�porn�mi �leny.
	Nech� d�le existuje $\lim_{n\to{}\infty}{a_{n}/b_{n}} = A \in{} \real^{*}$. Pak:

	\list{
		\listItem{\M{A \in{} (0, \infty) \Longrightarrow{}
			\sum_{n=1}^{\infty}{a_{n}} \textbox{konverguje} \Leftrightarrow{}
			\sum_{n=1}^{\infty}{b_{n}} \textbox{konverguje}}}
		\listItem{\M{A = 0 \Longrightarrow{}
			\sum_{n=1}^{\infty}{a_{n}} \textbox{konverguje} \Leftarrow{}
			\sum_{n=1}^{\infty}{b_{n}} \textbox{konverguje}}}
		\listItem{\M{A = \infty \Longrightarrow{}
			\sum_{n=1}^{\infty}{a_{n}} \textbox{konverguje} \Rightarrow{}
			\sum_{n=1}^{\infty}{b_{n}} \textbox{konverguje}}}
	}


	\proof{
	\list{
		\listItem{V�me, �e $\existss n_{0}\in{}\nat{}$ takov�, �e pro $\forall n\ge{}n_{0}$ plat�:

			$$ {1\over{}2}A \le{} {a_{n}\over{}b_{n}} \le{} 2A \qquad\textnote{jen �ikovn� epsilon} $$
			$$ {A\over{}2}b_{n} \le{} a_{n} \le{} 2Ab_{n} $$

			\iproofrightimpl{
			\M{\quad\sum_{n=1}^{\infty}{a_{n}} \textbox{konverguje}
				\becauseof{\ias{}{3}{}}{\Rightarrow}
				\sum_{n=1}^{\infty}{{A\over{}2}b_{n}} \textbox{konverguje}
				\becauseof{\ias{}{2}{}}{\Rightarrow}
				\sum_{n=1}^{\infty}{b_{n}} \textbox{konverguje}}
			}

			\iproofleftimpl{
			\M{\quad\sum_{n=1}^{\infty}{b_{n}} \textbox{konverguje}
				\becauseof{\ias{}{2}{}}{\Rightarrow}
				\sum_{n=1}^{\infty}{2Ab_{n}} \textbox{konverguje}
				\becauseof{\ias{}{3}{}}{\Rightarrow}
				\sum_{n=1}^{\infty}{a_{n}} \textbox{konverguje}}
			}
		}

		\listItem{Zvol $\eps=1$, pak $\existss{n_{0}}$ takov�, �e ${a_{n}/b_{n}} < 1$ pro $\forall{n\ge{}n_{0}}$, tedy $a_{n}<b_{n}$, \ias{}{3}{}.}
		\listItem{Zvol $\eps=1$, pak $\existss{n_{0}}$ takov�, �e ${a_{n}/b_{n}} > 1$ pro $\forall{n\ge{}n_{0}}$, tedy $a_{n}>b_{n}$, \ias{}{3}{}.}
	}
	\qed
	}


	\examples{
	\list{

	\listItem{\M{\sum_{n=1}^{\infty}{{n^3+1}\over{}\sqrt{3+n+2n^8}} \hypothetically{\approx{}} {n^3\over{}n^4} = {1\over{}n}}

		$$a_{n} = {{n^3+1}\over{}\sqrt{3+n+2n^8}}, \textbox{zvol} b_{n} = {1\over{}n}$$

		$$ \lim_{n\to{}\infty}{a_{n}\over{}b_{n}} = \lim_{n\to{}\infty}{{n^4+n}\over{}\sqrt{3+n+2n^8}} = {\sqrt{2}\over{}2} \in{} (0, \infty) $$

		Tedy dle (i) $\sum{a_{n}} \textbox{konverguje} \Leftrightarrow{} \sum{b_{n}}$ konverguje.
		My ale v�me, �e $\sum{b_{n}} = \sum{1\over{}n} = +\infty $ diverguje.

		$$ \sum_{n=1}^{\infty}{a_{n}} = +\infty \eqno\textnote{diverguje} $$
	}

\penalty-1000

	\listItem{\M{\sum_{n=1}^{\infty}{n^{15}\over{}3^n}}
		$a_{n} = {n^{15}\over{}3^n}$, zvol $b_{n} = {1/2^n}$

		Potom $$ \lim_{n\to{}\infty}{a_{n}\over{}b_{n}} = \lim_{n\to{}\infty}{n^{15}\over{}\left(3\over{}2\right)^n} = 0 $$ indukc�, Bernoulli, $\ldots$ (\DCV)

		Nav�c v�me, �e:
		$$ \sum{b_{n}} \textbox{konverguje} \textnote{geometrick� �ada} \becauseof{(ii)}{\Rightarrow} \sum{a_{n}} \textbox{konverguje} $$
	}

	}
	}

}




\theorem{5}{Cauchyovo odmocninov� krit�rium}{

	Nech� $\sum{a_{n}}$ je �ada s nez�porn�mi �leny:

	\list{
		\listItem{Nech� $\existss{q \in{} (0, 1)},\ \existss{n_{0} \in{} \nat{}},\ \forall{n \ge{} n_{0}}:\ \root n \of {a_{n}} \le{} q$. Pak $\sum{a_{n}}$ konverguje.}

		\listItem{Nech� $\limsup_{n\to{}\infty}{\root n \of{a_{n}}} < 1$. Pak $\sum{a_{n}}$ konverguje.}

		\listItem{Nech� $\lim_{n\to{}\infty}   {\root n \of{a_{n}}} < 1$. Pak $\sum{a_{n}}$ konverguje.}

		\listItem{Nech� $\limsup_{n\to{}\infty}{\root n \of{a_{n}}} > 1$. Pak $\sum{a_{n}}$ diverguje.}

		\listItem{Nech� $\lim_{n\to{}\infty}   {\root n \of{a_{n}}} > 1$. Pak $\sum{a_{n}}$ diverguje.}
	}


	\proof{
	\list{
		\listItem{Polo�me $b_{n} = q^n$, pak $a_{n} \le{} b_{n}$ pro $n \ge{} n_{0}$.
			D�le $\sum{b_{n}}$ konverguje (geometrick� �ada, $q < 1$)
			$\becauseof{\ias{}{3}{}}{\Rightarrow}$ tak�
			$\sum_{n=1}^{\infty}{a_{n}}$ konverguje.
		}

		\listItem{Ozna� $\limsup_{n\to{}\infty}\root n \of{a_{n}} = A < 1$. Zvol $\eps > 0$, $A + \eps < 1$.

			$$\existss{n_{0}},\ \forall{n \ge{} n_{0}}:\ \sup \{ \root n \of{a_{k}}, k \ge{} n \} \le{} A + \eps$$

			Tedy $\root n \of{a_{n}} \le{} A + \eps$, $n \ge{} n_{0}$.

			Ozna� $q \defined{} A + \eps < 1$, tvrzen� plyne z \ias{}{}{1}.
		}

		\listItem{Plyne z \ias{}{}{2}: $\existss{\rm lim} \Rightarrow{} \lim = \limsup$.}

		\listItem{
			$$ \limsup_{n\to{}\infty}\root n \of{a_{n}} > 1 $$
			  $$ \Rightarrow{} \existss{\textbox{vybran� posloupnost} \{a_{n_{k}}\}_{k=1}^{\infty},\ \root{\raise2pt\hbox{$\scriptscriptstyle n_k$}}\of{a_{n_{k}}} > 1} $$
			  $$ \Rightarrow{} a_{n_{k}} > 1\ \forall{k \in{} \nat{}} $$
			  $$ \Rightarrow{} \textbox{neplat� nutn� podm�nka} \lim_{n\to{}\infty}{a_{n}} = 0 \Rightarrow{} \sum\{a_{n}\} $$

			\scpart{Jin� pohled na posledn� krok}{
			   $$ \sum_{n=1}^{\infty}{a_{n}} \ge{} \sum_{n=1}^{\infty}{a_{n_{k}}} \textnote{nebo� $a_{n} \ge{} 0$} \ge{} \sum_{n=1}^{\infty}{1} = +\infty $$
			}
		}

		\listItem{Ihned plyne z \ias{}{}{4}.}
	}
		\qed
	}

}



\penalty-100

\theorem{6}{d'Alembertovo pod�lov� krit�rium}{
	Nech� $\sum{a_{n}}$ je �ada s kladn�mi �leny:

	\list{
		\listItem{$\existss{q \in{} (0, 1)}$, $\existss{n_{0} \in{} \nat{}}$, $\forall{n \ge{} n_{0}}$: ${a_{n+1}\over{}a_{n}} \le{} q \Rightarrow{} \sum{a_{n}} \textbox{konverguje}$.}

		\listItem{Nech� $\limsup_{n\to{}\infty}{a_{n+1}\over{}a_{n}} < 1$. Pak $\sum{a_{n}}$ konverguje.}

		\listItem{Nech� $\lim_{n\to{}\infty}{a_{n+1}\over{}a_{n}} < 1$. Pak $\sum{a_{n}}$ konverguje.}

		\listItem{Nech� $\lim_{n\to{}\infty}{a_{n+1}\over{}a_{n}} > 1$. Pak $\sum{a_{n}}$ diverguje.}
	}


	\proof{
	\list{
		\listItem{
			$$ a_{n_{0}+1} \le{} qa_{n_{0}} $$
			$$ a_{n_{0}+2} \le{} qa_{n_{0}+1} \le{} qa_{n_{0}} $$
			$$ a_{n_{0}+k} \le{} q^{k}a_{n_{0}} $$
			$$ \Rightarrow{} \sum_{n=1}^{\infty}{a_{n}} \le{} \undernote{\sum_{n=1}^{\infty}{a_{n}} + a_{n_{0}} \sum_{n=1}^{\infty}{q^k}}{konverguje} $$
			$$ \becauseof{\ias{}{3}{}}{\Rightarrow} \sum{a_{n}} \textbox{konverguje} $$
		}

		\listItem{(i) $\Rightarrow{}$ (ii) (\DCV)}
		\listItem{(ii) $\Rightarrow{}$ (iii) (\DCV)}
		\listItem{
			$$ \lim_{n\to{}\infty}{a_{n+1}\over{}a_{n}} > 1 \Rightarrow{} \existss{n_{0},\ \forall{n\Rightarrow{}n_{0}}}:\ {a_{n+1}\over{}a_{n}} > 1 $$
			$$ \Rightarrow{} \{a_{n}\} \textbox{je rostouc� od jist�ho indexu} $$
			$$ \Rightarrow{} \lim_{n\to{}\infty}{a_{n}} \not= 0 \becauseof{\ias{}{1}{}}{\Rightarrow} \sum{a_{n}} \textbox{diverguje} $$
		}
	}
	\qed
	}
}


\scpart{Pozn�mka o nebezpe��ch}{
\list{
	\listItem{$\lim_{n\to{}\infty}{\root n\of{a_{n}}} = 1 \Rightarrow{}$ nev�me nic.}

	\listItem{$\lim_{n\to{}\infty}{a_{n+1}\over{}a_{n}} = 1 \Rightarrow{}$ nev�me nic.}

	\example{$\sum{1\over{}n}$ diverguje, $\sum{1\over{}n^2}$ konverguje, ob� maj� limitu 1.}

	\listItem{Parametr $q < 1$ je d�le�it�!
		\example{
			 $$\root n\of{a_{n}} < 1 \not\Rightarrow{} \sum{a_{n}} \textbox{konverguje}\qquad\forall{n \ge n_0}$$
			 $${a_{n+1}\over{}a_{n}} < 1 \not\Rightarrow{} \sum{a_{n}} \textbox{konverguje}\qquad\forall{n \ge n_0}$$
		}
	}

	\penalty-100
	\listItem{Jestli�e \M{\limsup_{n\to{}\infty}{a_{n+1}\over{}a_{n}} > 1}, �ada divergovat nemus�.
		\example{
			 \M{1 + {1\over{}2} + {1\over{}4} + {1\over{}8} + \cdots + {1\over{}2^n}} konverguje.
			 Po zp�eh�zen� tak�, ale limes superior u� nesed�.
		}
	}
}
}


\example{
	Vy�et�ete, pro kter� $a \ge{} 0$ konverguje �ada \M{\sum_{n=1}^{\infty}{a^n n^n\over{}n!}}.

	\scpart{Pozorov�n�}{
		\list{
			\listItem{$a = 0$: $\sum{a_{n}}$ konverguje.}
			\listItem{$a$ moc velk�: $\sum{a_{n}}$ diverguje.}
			\listItem{$a = 1$: $\sum_{n=1}^{\infty}{n^n\over{}n!} = +\infty$
						$\Rightarrow{}$ $\sum{a_{n}}$ konverguje.}
		}
	}

	Dle \ref{d'Alembert}a:

	$$ {a_{n+1}\over{}a_{n}} = {{a^{n+1}(n+1)^{n+1}}\over{}(n+1)!} {n!\over{}{a^n n^n}} = a \left({{n+1}\over{}n}\right)^n = a \left(1 + {1\over{}n}\right)^n $$
	$$ \Rightarrow{} \lim_{n\to{}\infty}{a_{n+1}\over{}a_{n}} = a e $$
	$$ \Rightarrow{} a \in{} (0, {1/e}) \Rightarrow{} \sum{a_{n}} \textbox{konverguje} $$
	$$ \Rightarrow{} a \in{} ({1/e}, \infty) \Rightarrow{} \sum{a_{n}} \textbox{diverguje} $$
	$$ \Rightarrow{} a = {1\over{}e} \Rightarrow{} \textbox{nev�me nic} $$

	\scpart{Zb�vaj�c� vy�et�en�}{
		$$ a = {1\over{}e} \Rightarrow{} \sum_{n=1}^{\infty}\left(\left({n\over{}e}\right)^n {1\over{}n!}\right) $$

		Diverguje, nebo�
		$$\lim_{n\to{}\infty}\left(\left({n\over{}e}\right)^n {1\over{}n!}\right) \not= 0$$

		\DCV
	}

}


\theorem{7}{Cauchyovo kondenza�n� krit�rium}{
	Nech� $\sum{a_{n}}$ je �ada s nez�porn�mi �leny a nech�
	$\existss{n_{0}\in{}\nat{}}$ takov�, �e pro $\forall{n\ge{}n_{0}}$ plat� $a_{n+1}\le{}a_{n}$. Pak:
	$$ \sum_{n=1}^{\infty}{a_{n}} \textbox{konverguje} \Longleftrightarrow{}
		\sum_{n=1}^{\infty}{2^n a_{2^n}} \textbox{konverguje} $$


\penalty-1000

	\proof{
		
		\iproofleftimpl{P��mo:

			Nech� \M{\sum{2^n a_{2^n}}} konverguje. Ozna�:

			$$s_{n} = a_{1} + \cdots + a_{n}$$
			$$t_{k} = a_{1} + 2a_{2} + 4a_{4} + \cdots + 2^k a_{2^k}$$

			V�me: \M{\lim_{k\to{}\infty}{t_{k}} < \infty}

			Potom k dan�mu $n$ $\existss{k}$: $n < 2^k$.

			$$ s_{n} = a_{1} + a_{2} + a_{3} + a_{4} + \cdots + a_{n} $$
			$$ \le{} a_{1} + (a_{2} + a_{3}) + (a_{4} + \cdots + a_{7}) + \cdots + (a_{2^k} + \cdots + a_{2^{k+1}-1}) $$
			$$ \le{} a_{1} + 2a_{2} + 4a_{4} + \cdots + 2^k a_{2^k} = t_{k} $$
			$$ \Rightarrow{} \forall{n}:\quad s_{n} \le{} \lim_{t\to{}\infty}{t_{k}} \Rightarrow{} \textbox{$s_{n}$ je omezen�} \Rightarrow{} \sum{a_{n}} \textbox{konverguje} $$
		}

		\iproofrightimpl{Nep��mo:

			Nech� \M{\sum{2^n a_{2^n}}} diverguje.

			V�me: \M{\lim_{k\to{}\infty}{t_{k}} = +\infty}

			Potom k dan�mu $n$ $\existss{k}$: $n > 2^k$.

			$$ s_{n} = a_{1} + a_{2} + a_{3} + a_{4} + \cdots + a_{n} $$
			$$ = a_{1} + a_{2} + (a_{3} + a_{4}) + (a_{5} + \cdots + a_{7}) + \cdots + (a_{2^{k-1}+1} + \cdots + a_{2^k}) $$
			$$ \ge{} {a_{1}\over{}2} + a_{2} + 2a_{4} + 4a_{8} + \cdots + 2^{k-1} a_{2^k} = {1\over{}2} t_{k} $$
			$$ \Rightarrow{} \lim_{n\to{}\infty}{s_{n}} \ge{} \lim_{k\to{}\infty}{t_{k}} = +\infty $$
			$$ \Rightarrow{} \sum{a_{n}} \textbox{diverguje} $$
			\qed
		}

	}


	\example{
		Pro kter� $\a\in{}\real{}$ konverguje �ada \M{\sum_{n=1}^{\infty}{1\over{}n^\a}}?

		Z p�vodn�ch krit�ri� nevypl�v� nic, nebo�
		$\lim_{n\to{}\infty}{\root n\of{1\over{}n^\a}} = 1$,
		$\lim_{n\to{}\infty}{(n+1)^\a\over{}n^\a} = 1$.
		Nav�c v�me, �e �ada konverguje pro $\a = 2$ a diverguje pro $\a = 1$.
		Pro $\a \le{} 0$ diverguje tak�.
		Pro $\a > 0$ je $1\over{}n^\a$ klesaj�c�, tedy lze u��t \ias{}{7}{}.

		$$ \sum{1\over{}n^\a} \textbox{konverguje} \becauseof{\ias{}{7}{}}{\Longleftrightarrow}
			\sum_{n=1}^{\infty}\left(2^n {1\over{}(2^n)^\a}\right) \textbox{konverguje} $$
		   $$ \Leftrightarrow{} \sum_{n=1}^{\infty}{\left(1\over{}2^{\a-1}\right)^n} \textbox{konverguje} $$
		   $$ \Leftrightarrow{} \a > 1 \textnote{geometrick� �ada} $$
	}

	\scpart{Z�v�r}{
		$$ \sum_{n=1}^{\infty}{1\over{}n^\a} \textbox{konverguje} \Leftrightarrow{} \a > 1 $$
	}


	\example{
		$$ \sum_{n=2}^{\infty}{1\over{}n\log^\a n} \textbox{konverguje} \Leftrightarrow{} \a > 1 $$

		\proof{
			$$ \sum{} \textbox{konverguje} \Leftrightarrow{} \sum_{n=2}^{\infty}\left(2^n {1\over{}2^n\log^\a2^n}\right) \textbox{konverguje} \Leftrightarrow{} $$
			$$ \Leftrightarrow{} {1\over{}\log^\a2} \sum_{n=1}^{\infty}{1\over{}n^\a} \textbox{konverguje} \Leftrightarrow{} \a > 1 $$
		}
	}

}



\theorem{8}{Raabeovo kriterium}{
	Nech� $\sum{a_{n}}$ je �ada s kladn�mi �leny.

	$$\lim_{n\to{}\infty}{n\left({a_{n}\over{}a_{n+1}} - 1\right)} > 1 \Rightarrow{} \sum{a_{n}} \textbox{konverguje}$$
	$$\lim_{n\to{}\infty}{n\left({a_{n}\over{}a_{n+1}} - 1\right)} < 1 \Rightarrow{} \sum{a_{n}} \textbox{diverguje}$$

	\example{
		 $$ \sum{a_{n}},\ a_{n} = \left({{1\cdot4\cdot7\cdots(3n-2)}\over{}{3\cdot6\cdot9\cdots3n}}\right)^2 $$

		 \DCV
	}

}



}

\bend
