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A problem asks me to determine if the series

$$\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}$$

converges or diverges.

(from the textbook Calculus by Laura Taalman and Peter Kohn (2014 edition); section 7.6, p. 639, problem 33)

I am allowed to use the ratio test first and then any other convergence/divergence test if the former test does not work. In my original work, I attempted the ratio test and it was rendered inconclusive.

$$ a_k = \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)} $$

$$ a_{k + 1} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2(k+1))}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2(k+1)-1)} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)} $$

$$ \frac{a_{k + 1}}{a_k} = \frac{\frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)}}{\frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)} \times \frac{1 \times 3 \times 5 \times \cdots \times (2k-1)}{2 \times 4 \times 6 \times \cdots \times (2k)} = \frac{2k+2}{2k+1}$$

Evaluating $\rho = \lim_{x \to \infty} \frac{a_{k + 1}}{a_k}$ will determine if $\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}$ converges or diverges. The conclusions for the ratio test are as follows:

$\circ$ If $\rho < 1$, then $\sum_{k=1}^\infty a_k$ converges.

$\circ$ If $\rho > 1$, then $\sum_{k=1}^\infty a_k$ diverges.

$\circ$ If $\rho = 1$, then the test is inconclusive.

$$ \rho = \lim_{x \to \infty} \frac{a_{k + 1}}{a_k} = \lim_{x \to \infty} \frac{2k+2}{2k+1} = 1$$

Since $\rho = 1$, the ratio test is rendered inconclusive, as I stated earlier. I will have to use other convergence/divergence tests to solve the problem.

My issue is that I'm not sure which other convergence/divergence test to use. Any suggestions?

Many thanks for the help.

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The series is divergent since $a_k>1$ for every $k$.

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    $\begingroup$ @Ksquared in order for your sum to converge you need that $(a_k) \to 0$ $\endgroup$ – ÍgjøgnumMeg Sep 15 '16 at 18:23
  • $\begingroup$ @Ed_4434 I see. $\endgroup$ – Ksquared Sep 15 '16 at 18:30
  • $\begingroup$ @OlivierMoschetta What test did you use? $\endgroup$ – Ksquared Sep 15 '16 at 18:38
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    $\begingroup$ @Ksquared I think the point is that $2k > 2k-1$ for all $k \geq 1$, so each term of the series is larger than $1$ and then you have an infinite sum of terms that are larger than 1. $\endgroup$ – ÍgjøgnumMeg Sep 15 '16 at 18:46
  • $\begingroup$ @ksquared My first comment is actually false. It should be the other way around, that is, if $\sum_{n=1}^\infty a_n$ is convergent, THEN $(a_n) \to 0$. A counter-example to my first comment is the harmonic series. $\endgroup$ – ÍgjøgnumMeg Sep 15 '16 at 20:54
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By Stirling approximation, we have that $$a_k=\frac{2\times 4\times 6\times\cdots\times(2k)}{1\times 3\times 5\times\cdots\times(2k-1)}=\frac{4^k\cdot (k!)^2}{(2k)!}=\frac{4^k}{\binom{2k}{k}}\sim \sqrt{\pi k}.$$ Hence the series $\sum_{k\geq 1} a_k$ diverges.

More generally the series $\sum_{k\geq 1} \frac{a_k}{k^r}$ converges iff $r-1/2>1$, that is $r>3/2$.

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$$\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}=$$

$$\sum_{k=1}^\infty \frac{\prod_{j=1}^k(2j)}{\prod_{j=1}^k(2j-1)}=$$

$$\sum_{k=1}^\infty \prod_{j=1}^k\frac{(2j)}{(2j-1)}$$

The product $a_k = \prod_{j=1}^k\frac{(2j)}{(2j-1)}$ is $$\left(1+\dfrac11\right)\left(1+\dfrac13\right)\left(1+\dfrac15\right)\cdots \left(1+\dfrac1{2n-1}\right)$$ An infinite product $\lim_{k \to \infty} \left(1+a_k\right)$ converges to a non-zero number if one of the $\sum_{k \to \infty} \vert a_k \vert$ converges. Conclude what you want from this.

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