The series $\sum a_n$ converges, where $a_n$ is the product of fractions from $1/2$ to $(2n-3)/(2n-2)$, divided by $2n-1$

Question

Prove the series given by the sequence

$$a_n= \frac{1}{2}·\frac{3}{4}·\ldots ·\frac{2n-3}{2n-2}·\frac{1}{2n-1}$$

converges

The series is

$$\sum_{n=1}^\infty a_n = 1+\frac{1}{2}·\frac{1}{3}+\frac{1}{2}·\frac{3}{4}·\frac{1}{5}+\frac{1}{2}·\frac{3}{4}·\frac{5}{6}·\frac{1}{7} + \cdots$$

PS: The ratio test does not help.

I've edited your question using LaTeX to make it more readable. This page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good. — Antonio Vargas, Jul 31 '14 at 13:42
In full generality, do you think the root test can be conclusive when the ratio test is not because the limit of the ratios exists but is 1? — Did, Jul 31 '14 at 13:44

Did · Answer 1 · 2014-07-31T15:24:22.223

3

Exercise: Assume that $a_n\gt0$ for every $n$ and that $a_{n+1}/a_n\to1$. Show that $(a_n)^{1/n}\to1$.

Consequence: If the ratio test is not conclusive because the limit of the ratios exists and is $1$, then the root test is not conclusive either.

edited Jul 31 '14 at 15:24

answered Jul 31 '14 at 13:46

Did

284,245

we can use that lim sup and lim inf inequality but by root test the limit has to be less than 1 – UNM Jul 31 '14 at 13:50
1

Sorry but what are you trying to say? – Did Jul 31 '14 at 13:50
Sorry yeah you are correct I thought you were giving a proof. – UNM Jul 31 '14 at 13:54

score 1 · Answer 2 · answered Jul 31 '14 at 14:49

If $a_k = \frac{1}{(2k-1)}\prod_{k=2}^{n}\frac{(2k-3)}{(2k-2)}$ then consider $v_k = e^{\log a_k}$. After a bit of algebra you get tonnes of cancellations, i.e. an expression of the form $\log (\frac{2}{1} \frac{3}{2} \ldots \frac{2n-2}{2n-3}\big) = \log (2n-2)$, and the final expression (don't forget the extra term in the denominator!) is of the form $\sum_{n=1}^{\infty} \frac{1}{(2n-2)(2n-1)}$ which you can compare to a famous series and conclude divergence.

Except that the series converges. – Did Jul 31 '14 at 15:26 — Did, Jul 31 '14 at 15:26

The series $\sum a_n$ converges, where $a_n$ is the product of fractions from $1/2$ to $(2n-3)/(2n-2)$, divided by $2n-1$

2 Answers2