Prove the series defined by P(n) = (1 *3 * 5 * (2n-1))/(2*4*6 * (2n)) is convergent
It is monotone decreasing and bounded below by zero, but is that enough to say?
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$\begingroup$Note that
$$\displaystyle \ln \Pi_{n=1}^N \frac{2n-1}{2n}=\sum_{n=1}^N \ln \left(1-\frac{1}{2n}\right)\le -\frac12 \sum_{n=1}^N\frac 1n,$$ where we have used the known inequality $\ln(1+x)\le x,$ which holds for any $x>-1.$ Now, $s_N=\sum_{n=1}^N \frac 1n\to \infty$ as $N\to \infty.$ Thus $$\displaystyle \ln \Pi_{n=1}^N \frac{2n-1}{2n}\le -\frac12 s_N,$$ or equivalently,
$$\displaystyle \Pi_{n=1}^N \frac{2n-1}{2n}\le e^{-1/2s_N}.$$ Since the product is nonnegative and $-1/2s_n\to -\infty$ we conclude that
$$\displaystyle \Pi_{n=1}^N \frac{2n-1}{2n}\to 0.$$
$\endgroup$ 2 $\begingroup$The sequence $P(n)$ is convergent, the series $\sum P(n)$ is divergent.
From inequality of arithmetic and geometric means:$$\sqrt{1\cdot 3}\leq 2,\ \sqrt{3\cdot 5}\leq 4,\ldots, \sqrt{(2n-3)(2n-1)}\leq 2n-2.$$Multiplying them we obtain$$1\cdot 3\cdots (2n-3)\sqrt{2n-1}\leq 2\cdot 4\cdots (2n-2)$$. Therefore$$P(n)\leq \frac{\sqrt{2n-1}}{2n}.$$This shows that $P(n)\to 0$. On the other hand, similarly (using $\sqrt{2\cdot 4}\leq 3$, etc.), one can show that$$P(n)\geq \frac 1{2\sqrt n}.$$This shows that$$\sum_{n=1}^\infty P(n)=\infty.$$
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