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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 \(\lim _{nβ†’βˆž} z_ n=0\) but

\[ βˆ‘_{n=1}^∞ z_ n=∞ \quad . \]

Solution 1

[213]

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Authors: Ricci, Fulvio ; Pacini, Tommaso ; "Mennucci , Andrea C. G." .
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