# convergence tests for series $p_n=\frac{1\cdot 3\cdot 5…(2n-1)}{2\cdot 4\cdot 6…(2n)}$

If the sequence:

$p_n=\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n)}$

Prove that the sequence

$((n+1/2)p_n^2)^{n=\infty}_{1}$ is decreasing.

and that the series $(np_n^2)^{n=\infty}_{1}$ is convergent.

Any hints/ answers would be great.

I'm unsure where to begin.

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Do you mean series or sequence here? In a series you consider the sum all the terms, and it really does not make sense to say that a series of positive terms is decreasing. – Per Manne Nov 7 '12 at 12:05
sequence it is then, thanks for correction – redrum Nov 7 '12 at 12:27

Firstly, it should be noted that $\dfrac{p_{n+1}}{p_n}=\dfrac{2n+1}{2n+2}=\dfrac{n+\frac{1}{2}}{n+1},$ therefore, $$\frac{a_{n+1}}{a_n}=\frac{\left(n+1+\frac{1}{2} \right)p_{n+1}^2}{\left(n+\frac{1}{2} \right)p_{n}^2}=\frac{\left(n+\frac{3}{2} \right)\left(n+\frac{1}{2} \right)^2}{\left(n+\frac{1}{2} \right)\left(n+1 \right)^2}=\frac{\left(n+\frac{3}{2} \right)\left(n+\frac{1}{2} \right)}{\left(n+1 \right)^2}=\frac{n^2+2n+\frac{3}{4}}{n^2+2n+1}<1.$$ Next, $p_n$ can be rewritten as $$p_n=\frac{3}{2}\cdot \frac{5}{4}\cdot \ldots \cdot \frac{2n-1}{2n-2}\cdot \frac{1}{2n} > \frac{3}{2}\cdot \frac{1}{2n}=\frac{3}{4n},$$ which implies $$p_n^2>\frac{9}{16}\cdot\frac{1}{n^2}$$ and $$np_n^2>n\cdot\dfrac{9}{16}\cdot\dfrac{1}{n^2}=\dfrac{9}{16}\cdot\dfrac{1}{n},$$
so $\sum\limits_{n=1}^{\infty}{np_n^2}$ diverges.

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Hint 1: Show that (n+1/2)>=(n+1.5)(2n+1/2n+2)^2 for all positive integers n, then use induction to show that the first sequence is decreasing Hint 2: show that 1/2n<=p(n), thus 1/2n<= np(n)^2 therefore the second series diverges

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Hint 2: show that 1/2n<=p(n), thus 1/2n<= np(n)^2 therefore the second series diverges – Amr Nov 7 '12 at 12:22
Hint 3: To show that show that 1/2n<=p(n) you can use induction or write 1/n as (1/2)(2/3)(3/4)...(n-1/n) then show that each term in the previous product is less than or equal the corresponding term in the product (3/4)(5/6)...(2n-3/2n-2)(2n-1/2n) – Amr Nov 7 '12 at 12:25
are you sure the second series diverges, my question paper actually asks me to prove that it converges...? – redrum Nov 7 '12 at 12:31
by 'second series' I assume you meant $(np_n^2)^{n=\infty}_{1}$ – redrum Nov 7 '12 at 12:33
yes I meant the series 1p(1)^2+2p(2)^2+3p(3)^2+... – Amr Nov 7 '12 at 12:52