\subsubsection {}
Rozhodn�te, kter� z n�sleduj�c�ch �ad konverguj�.
$$\sum_{n=1}^{\infty}{1\over n}, \sum_{n=1}^{\infty}{1\over
\sqrt n}, \sum_{n=2}^{\infty}{1\over \sqrt{n(n-1)}}.$$
 



\subsubsection {}
Se�t�te �ady
$$\sum_{n=2}^{\infty}{1\over n^2-1} \text{ a
}\sum_{n+1}^{\infty}{1\over n(n+1)(n+2)(n+3)}.$$
 



\subsubsection {}
Rozhodn�te, kter� z n�sleduj�c�ch �ad konverguj�:
$$\sum_{n=0}^{\infty}(-1)^n{2n^2+3n+4\over 2n^2+5},
\sum_{n=20}^{\infty}(-1)^n{2n^2+3n+4\over (2n^2+5)^2},$$
$$\sum_{n=2}^{\infty}(-1)^n({n-1\over n+1})^n,
\sum_{n=0}^{\infty}{n^3\over n^3+1}.$$
 



\subsubsection {}
Rozhodn�te, zda konverguj� posloupnosti:
$$a_n=\prod_{n=1}^n{k+900\over 2k-1}, a_n=(1+{1\over
2})\cdot\dots\cdot(1+{1\over 2^n}).$$
 



\subsubsection {}
Rozhodn�te o konvergenci �ad:
$$\sum_{n=0}^{\infty}{\sin (nx)\over 2^n},
\sum_{n=0}^{\infty}{3+(-1)^n\over 2^{n+1}},$$
$$\sum_{n=2}^{\infty}{1\over 2n+1},
\sum_{n=0}^{\infty}{1\over (2n+1)^2}.$$
Sou�et posledn� z nich vyj�d�ete pomoc�
$\sum_{n=1}^{\infty}{1\over n^2}$. 



\subsubsection {}
Pro jak� posloupnosti $\epsilon_i$ spl�uj�c�
$|\epsilon_i|=1$ pro $i=1,\dots$ konverguje absolutn� �ada
$${\epsilon_1\over 2}+{\epsilon_2\over 3}+{\epsilon_3\over
2^2}+{\epsilon_4\over 3^2}+{\epsilon_5\over
2^3}+{\epsilon_6\over 3^3}+{\epsilon_7\over
2^4}+{\epsilon_8\over 3^4}+ \dots.$$
 



\subsubsection {}
Rozhodn�te o konvergenci �ad
$$\sum_{n=1}^{\infty}{n^2\over (2+{1\over n})^n},
\sum_{n=0}^{\infty}{n^5\over 2^n+3^n},
\sum_{n=1}^{\infty}{n\over 2^n},$$
$$\sum_{n=0}^{\infty}{1\over \sqrt{2n+1}\sqrt{2n+3}},
\sum_{n=1}^{\infty}{n^{n+{1\over n}}\over n+{1\over n}}.$$
 



\subsubsection {}
Pro kter� re�ln� $x$ konverguj� �ady
$$\sum_{n=1}^{\infty}{x^n\over n},
\sum_{n=0}^{\infty}x^n{n^2\over n^3+1}.$$
 



\subsubsection {}
Doka�te, �e konverguj� �ady
$$\sum_{n=0}^{\infty}{1\over (n+1)\sqrt{n+2}},
\sum_{n=1}^{\infty}{1\over n^{\alpha}} \text{ pro }\alpha>1.$$
 



\subsubsection {}
Doka�te postupn�
$$(a) e=\lim_{n\to\infty}(1+1+{1\over 2!}+{1\over
3!}+\dots+{1\over n!});$$
$$(b) e=1+1+{1\over 2!}+\dots+{1\over
n!}+{\alpha_n\over{n!n}}, \text{ kde }\alpha_n\in (0,1),$$
co� ��k�, jak dob�e $\sum_{k=0}^{\infty}{1\over k!}$
aproximuje $e$;\hfill\break
$$(c) e\text{ nen� racion�ln� ��slo.}$$
