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.


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

  • 1
    $\begingroup$ @Ksquared in order for your sum to converge you need that $(a_k) \to 0$ $\endgroup$ – Edward Evans 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
  • 1
    $\begingroup$ @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. $\endgroup$ – Edward Evans 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. $\endgroup$ – Edward Evans Sep 15 '16 at 20:54

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 $r>3/2$.


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

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

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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.