1
$\begingroup$

$$\sum_{n=1}^{\infty}\frac{1\cdot 3\cdot 5\cdot ...\cdot (2n-1)}{1\cdot 4\cdot 7\cdot ...(3n-2)}$$

Using Ratio test: $$\lim_{n\rightarrow \infty}\frac{\frac{2(n+1)-1}{3(n+1)-2}}{\frac{2n-1}{3n-2}}$$

which equals to : $$\lim_{n \to \infty}\frac{6n^{2}-n-2}{6n^{2}-n-1}$$ the result for latter is q=1. seemingly this is unclear if divergent or convergent.

The answer for this is convergent....

$\endgroup$
  • $\begingroup$ No, when $n$ increases, there are more factors. Your ratio is incorrect. $\endgroup$ – Yves Daoust Oct 9 '18 at 13:54
3
$\begingroup$

List out your $a_n$ clearly.$$a_n =\prod_{i=1}^n \left(\frac{2i-1}{3i-2}\right)$$

$$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=\lim_{n \to \infty}\frac{2(n+1)-1}{3(n+1)-2}=\frac{2}{3}$$

$\endgroup$
1
$\begingroup$

Hint:

$$\frac{1\cdot3\cdot5\cdots\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=\frac{1\cdot3\cdot5\cdots\cdot(2n-1)\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=2n+1$$

and not

$$\frac{1\cdot3\cdot5\cdots\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=\frac{2n+1}{2n-1}.$$

$\endgroup$
1
$\begingroup$

The ratio test will work if done correctly $$ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \frac { \left(\! {\large{\frac {1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)\,\cdot\,(2n+1)\!} {1\,\cdot\,4\,\cdot\,7\,\cdots\,(3n-2)\,\cdot\,(3n+1)\!} }} \! \right) } {\left(\!\large{{\frac {1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)} {1\,\cdot\,4\,\cdot\,7\,\cdots\,(3n-2)} }}\!\!\right)} = \lim_{n\to\infty} \frac{2n+1}{3n+1} = \frac{2}{3} $$ But here's another way . . .

Show that the sequence of fractions $$\frac{1}{1},\frac{3}{4},\frac{5}{7},...,\frac{2n-1}{3n-2}$$ is strictly decreasing.

From that, it follows that the $n$-th term is bounded above by $\bigl({\large{\frac{3}{4}}}\bigr)^{n-1}$.

Finish via a comparison test.

$\endgroup$
0
$\begingroup$

$$a_k=\prod_{k=1}^{n}\frac{2k-1}{3k-2}=\left(\frac{2}{3}\right)^n\cdot\frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\cdot B\left(n+\tfrac{1}{2},\tfrac{1}{6}\right)\tag{1}$$ implies $$\begin{eqnarray*} \sum_{k\geq 1}a_k &=& \frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\int_{0}^{1}\sum_{n\geq 1}\left(\frac{2}{3}\right)^n (1-x)^{-5/6} x^{n-1/2}\,dx \\&=&\frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\int_{0}^{\pi/2}\frac{4(1-\cos^2\theta)}{\cos^{2/3}\theta(1+2\cos^2\theta)}\,d\theta\\&=&\frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\int_{0}^{1}12\sqrt{1-z^6}\,\frac{dz}{1+2z^6}\tag{2}\end{eqnarray*}$$ hence trivially $$ \sum_{k\geq 1}a_k = \tfrac{1}{2}\cdot\phantom{}_2 F_1\left(1,\tfrac{3}{2};\tfrac{5}{3};\tfrac{2}{3}\right) \leq \frac{3}{\pi}2^{4/3}\,\Gamma\left(\tfrac{2}{3}\right)^2.\tag{3}$$ In general, $\phantom{}_{p+1}F_p(\ldots; z)$ is convergent for any $|z|<1$.

$\endgroup$

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.