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Does the series

$$\sum_{k=1}^\infty\left(\prod_{j=1}^k\frac{2j-1}{2j+2}\right)$$ converge?

I am supposed to use the theorem below:

Theorem: let $x_k>0 \forall k \in \mathbb{N}$. If there exists $r<1 $ such that $$\frac{x_{k+1}}{x_{k}}<r$$ when $k\geq k_0\in\mathbb{N}$, then $\sum_{k=1}^{\infty} x_k$ converges.

What I have tried:

$$\frac{x_{k+1}}{x_{k}}=\frac{\frac{1}{4}\cdot{}\frac{3}{6}\cdot{}\frac{5}{8}\cdot{}\dots \cdot(2k+1)}{\frac{1}{4}\cdot{}\frac{3}{6}\cdot{}\frac{5}{8}\cdot{}\dots \cdot(2k+4)}=\frac{(2k+1)}{(2k+4)}=\frac{(1+\frac{1}{2k})}{(1+\frac{2}{k})}$$this is where I got stuck.

Arnaud Mortier
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1likeMaths
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  • The ratio test failed you. Are you sure this is the only one you can use? – Yuriy S Feb 05 '18 at 16:26
  • Yes, I am supposed to use the theorem stated above. – 1likeMaths Feb 05 '18 at 16:29
  • The ratio test can't give any conclusion on the convergence if the limit of the ratio for $k \to \infty$ is $1$, which seems to be the case here. So, using this method, your answer would be "there's no way to know if the series converges or not" – Yuriy S Feb 05 '18 at 16:34
  • @YuriyS how did you find that the series diverges? I think that it doesn't - see below. – Arnaud Mortier Feb 05 '18 at 16:46
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    @1likeMaths the lest two terms in your huge fraction should be $\dfrac{2k+1}{2k+4}$, shouldn't they? – Arnaud Mortier Feb 05 '18 at 16:47
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    @ArnaudMortier, I have made a mistake when writing the series, so my comment was wrong. Thank you – Yuriy S Feb 05 '18 at 16:51

2 Answers2

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An extension of the ratio test, called Raabe's test, states that in case the ratio test is inconclusive because the limit is $1$, if $$\lim_{k\to \infty} k\left(\left|\dfrac{a_k}{a_{k+1}}\right|-1\right)=R>1$$ then the series absolutely converges.

In your case, $R=\dfrac{3}{2}>1.$

Note that the fraction here is $\dfrac{a_k}{a_{k+1}}$ instead of $\dfrac{a_{k+1}}{a_{k}}$.

Arnaud Mortier
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As already remarked, it is convergent by Raabe's test. As an alternative, one may use Gautschi's inequality and the $p$-test. Additionally the value of the series can be computed through Euler's Beta function:

$$ \sum_{k\geq 1}\frac{\Gamma\left(k+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\,\Gamma(k+2)}=\frac{2}{\pi}\sum_{k\geq 1}B\left(k+\tfrac{1}{2},\tfrac{3}{2}\right)=\frac{2}{\pi}\int_{0}^{1}\sum_{k\geq 1}x^{k-1/2}\sqrt{1-x}\,dx\tag{1}$$ and the RHS equals: $$ \frac{2}{\pi}\int_{0}^{1}\sqrt{\frac{x}{1-x}}\,dx =\frac{2}{\pi}B\left(\tfrac{3}{2},\tfrac{1}{2}\right)=\color{red}{\large 1}.\tag{2}$$

Jack D'Aurizio
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