# Improper use of Test of Divergence

Determine if the series $$\sum_{n=1}^{\infty}\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot 2n-1}{2 \cdot 5 \cdot 8 \cdot \cdots \cdot 3n-1}$$ converges or diverges.

I applied the Test of Divergence, so $\lim_{n \rightarrow \infty}\frac{2n-1}{3n-1} = \frac{2}{3} \ne 0$, which implies the series is divergent. However, the correct answer suggested the use of the ratio test, where it concluded that $\frac{2}{3} < 1$, which implies the series is convergent.

Therefore, can we not use the Test of Divergence to test this series for convergence, and if not, why is that the case? Any help would be appreciated!

• Why are you taking the limit of ${2n-1\over 3n-1}$ in your attempt to do the test for divergence? That test requires you to take the limit of the $n^{th}$ term, and that's not what the $n^{th}$ term is. – Adam Hughes Apr 1 '17 at 17:48
• You forgot about all the numbers before $2n-1$ and $3n-1$. We have $\sum a_n$ but $a_n \neq \frac{2n-1}{3n-1}$ – Ahmed S. Attaalla Apr 1 '17 at 17:48

The test of divergence should be applied to each term as a whole: we have $$\lim_{n\to\infty} a_n = \lim_{n \to \infty} \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)}{2 \cdot 5 \cdot 8 \cdot \cdots \cdot (3n-1)} = 0$$ and therefore $\sum_{n=1}^\infty a_n$ is not necessarily divergent.

It's not immediately obvious why this limit is $0$: for instance, you could say that the numerator is less than $$2 \cdot 4 \cdot 6 \cdot \dots \cdot 2n = 2^n \cdot n!$$ and the denominator is greater than $$1 \cdot 3 \cdot 6 \cdot \cdots \cdots (3n-3) = 3^{n-1} \cdot (n-1)!$$ so $$\lim_{n \to \infty} \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)}{2 \cdot 5 \cdot 8 \cdot \cdots \cdot (3n-1)} \le \lim_{n\to\infty} \frac{2^n \cdot n!}{3^{n-1} \cdot (n-1)!} = \lim_{n\to \infty} 3n \cdot \left(\frac23\right)^n = 0.$$ But at the very least, you should admit that the limit is not the same as $\lim_{n\to\infty} \frac{2n-1}{3n-1} = \frac23$.

It is convergent and we may also compute its sum:

$$\sum_{n\geq 1}\frac{(2n-1)!!}{(3n-1)!!!} = \frac{\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}\sum_{n\geq 1}\left(\frac{2}{3}\right)^n B\left(n+\frac{1}{2},\frac{1}{6}\right)$$ equals: $$\small \frac{\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}\int_{0}^{1}\sum_{n\geq 1}\left(\frac{2}{3}\right)^n (1-x)^{-5/6} x^{n-1/2}\,dx =\frac{\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}\int_{0}^{1}\frac{2\sqrt{x}}{(1-x)^{5/6}(3-2x)}\,dx.$$

This proves the hypergeometric identity

$$\phantom{}_2 F_1\left(1,\frac{3}{2};\frac{5}{3};\frac{2}{3}\right)=-12 - 5\cdot\phantom{}_2 F_1\left(-\frac{1}{2},1;\frac{2}{3};\frac{2}{3}\right)+7\cdot \phantom{}_2 F_1\left(\frac{1}{2},1;\frac{2}{3};\frac{2}{3}\right).$$

By Ratio test it converges since

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{2n+1}{3n+2}=\frac23<1$$