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Use Gauss’ test to prove the convergence or divergence of a given series

Consider the series

    \[ \sum_{n=1}^{\infty} \left( \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \right)^k. \]

Use Gauss’ test (from the previous exercise, Section 10.16, Exercise #17) to prove that the series converges for k > 2 and diverges for k \leq 2.


Proof. From the definition of the series we have the nth and n+1st terms,

    \[ a_n = \let( \frac{1 \cdot 3 \cdot 5 \cdots (2n+1)}{2 \cdot 4 \cdot 6 \cdots (2n+2)} \right)^k, \qquad a_{n+1} = \left( \frac{1 \cdot 3 \cdot 5 \cdots (2n+1)}{2 \cdot 4 \cdot 6 \cdots (2n+2)}\right)^k. \]

Therefore,

    \begin{align*}  \frac{a_{n+1}}{a_n} &= \left( \frac{1 \cdot 3 \cdot 5 \cdots (2n+1)}{2 \cdot 4 \cdot 6 \cdots (2n+2)} \right)^k \cdot \left( \frac{2 \cdot 4 \cdot 6 \cdots (2n)}{1 \cdot 3 \cdot 5 \cdots (2n-1)} \right)^k \\[9pt]  &= \left( \frac{2n+1}{2n+2} \right)^k \\[9pt]  &= \left( 1 - \frac{1}{2(n+1)}\right)^k. \end{align*}

Using the Taylor expansion,

    \[ (1-x)^k = 1 - kx + \frac{1}{2}k(k-1)x^2 - \cdots. \]

So we have,

    \begin{align*}  \left( 1 - \frac{1}{2(n+1)} \right)^k &= 1 - \frac{k}{2}\left( \frac{1}{n+1} \right) + \frac{k(k-1)}{8} \left( \frac{1}{(n+1)^2} \right) - \cdots \\[9pt]  &= 1 - \frac{k/2}{n+1} + \frac{f(n)}{n^2}, \end{align*}

where f(n) is bounded. Letting A = \frac{k}{2} we apply Gauss’ test to conclude that \sum a_n converges if k > 2 and diverges if k \leq 2. \qquad \blacksquare

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