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Exercise
28
[20Z]Note:Task of 26 Jan 2016. (Solved on 2022-01-20)
Let
\[ z_ n = \frac{1β
3β
5β
7 \cdots (2n-1)}{2β
4β
6β
8 \cdots ~ (2n)} \quad ; \]
Show that \(\lim _{nββ} z_ n=0\) but
\[ β_{n=1}^β z_ n=β \quad . \]
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