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[tex]\lim_{n \to \infty} (\frac{1}{1\cdot 3} + \frac{1}{3\cdot 5} + \frac{1}{5 \cdot 7} - ... +\frac{1}{(2n-1)(2n+1)})[/tex]
Skorzystajmy z własności, że [tex]\frac{1}{n(n+2)} = \frac{\frac{1}{2}}{n} - \frac{\frac{1}{2}}{n+2}[/tex]:
[tex]\lim_{n \to \infty} (\frac{\frac{1}{2}}{1} - \frac{\frac{1}{2}}{3} + \frac{\frac{1}{2}}{3}} - \frac{\frac{1}{2}}{5} + \frac{\frac{1}{2}}{5} -\frac{\frac{1}{2}}{7} + ... +\frac{\frac{1}{2}}{2n-1} -\frac{\frac{1}{2}}{2n+1}) = \\\\\\ = \lim_{n \to \infty} (\frac{1}{2} - \frac{\frac{1}{2}}{2n+1})[/tex]
Sprowadźmy do wspólnego mianownika:
[tex]\lim_{n \to \infty} (\frac{1}{2} -\frac{1}{4n+2}) = \lim_{n \to \infty} (\frac{2n+1}{4n+2} - \frac{1}{4n+2}) = \lim_{n \to \infty} (\frac{2n}{4n+2}) = \frac{2}{4} = \frac{1}{2}[/tex]
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