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Infinite Serieshard
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Determine the convergence of ∑n=1∞1⋅3⋅5…(2n−1)2⋅4⋅6…(2n)⋅12n+1\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{2 \cdot 4 \cdot 6 \dots (2n)} \cdot \frac{1}{2n+1}∑n=1∞​2⋅4⋅6…(2n)1⋅3⋅5…(2n−1)​⋅2n+11​.