Convergence of the Series - $\sum_{n=1}^{\infty} \frac{\prod_{k=1}^{n}{(3k-1)}}{\prod_{k=1}^{n}(4k-3)}$

Question

Prove that the following series is convergent.

$$\sum_{n=1}^{\infty} \frac{\prod_{k=1}^{n}{(3k-1)}}{\prod_{k=1}^{n}(4k-3)}$$

I don't know for where to begin.

André Nicolas · Accepted Answer · 2013-06-09T23:21:51.273

4

Hint: The Ratio Test is perfect for this. If $a_n$ is the $n$-th term, then $$\frac{a_{n+1}}{a_n}=\frac{3n+2}{4n+1}.$$

Detail: Note that $$a_{n+1}=\frac{\prod_1^{n+1} (3k-1)}{\prod_1^{n+1} (4k-3)}.$$ But $$\prod_{k=1}^{n+1}(3k-1)=\left(\prod_1^n (3k-1)\right)\left(3(n+1)-1\right),$$ and $$\prod_{k=1}^{n+1}(4k-3)=\left(\prod_1^n (4k-3)\right)\left(4(n+1)-3\right).$$ Now when we calculate $\dfrac{a_{n+1}}{a_n}$, we see there is very nice cancellation.

edited Jun 09 '13 at 23:21

answered Jun 09 '13 at 23:02

André Nicolas

507,029

I applied the ratio test but didn't find this result. My ratio test gives me:
$$\frac{a_{n+1}}{a_n} = \frac{\prod{3k+2}.\prod{4k-3}}{\prod{4k+3}.\prod{3k-1}}$$
– Otávio Rapôso Jun 09 '13 at 23:09
I will add the calculation. You are having trouble with the product notation. – André Nicolas Jun 09 '13 at 23:13
Oh, I understood now. I just need to add one extra factor. Thank you. – Otávio Rapôso Jun 09 '13 at 23:19

score 1 · Answer 2 · answered Jun 09 '13 at 23:02

1

Hint: You can use the ratio test.

answered Jun 09 '13 at 23:02

Mhenni Benghorbal

47,431

Convergence of the Series - $\sum_{n=1}^{\infty} \frac{\prod_{k=1}^{n}{(3k-1)}}{\prod_{k=1}^{n}(4k-3)}$

2 Answers2