1
$\begingroup$

$$\sum\limits_{n=1}^{\infty}(-1)^{n+1}\frac{3^{n}}{3 \cdot 5 \cdot 7 \cdots (2n+1)}.$$

I am trying to use Leibniz in order to prove that the series converges. I don't know if I am doing it correctly. Here it is.

I want to prove that it is decreasing.

$$\frac{a_{n+1}}{a_{n}}= \frac{3}{2n+3}<1 ,$$ so it is decreasing.

Then we want the $\lim\limits_{n \to \infty}a_{n}=0$. The problem is that the limit is not zero. So either I am missing something or the series does not converge.

  • 3
    $a_n$ *does* tend to $0$.2011-10-29
  • 1
    You can use the ratio test to show absolute convergence. So, the terms have to go to $0$.2011-10-29
  • 1
    To add to @Joe's comment, once you show absolute convergence (through ratio test), convergence of the series follows automatically. No need to invoke Leibniz in that case. // Alternately, show that the terms $a_n$ are bounded by the geometric series $a_1 (3/5)^{n-1}$ (where $a_1$ is the first term of $a$), which converges to $0$.2011-10-29
  • 0
    Can you explain why $a_{n}$ tends to $0$?2011-10-29
  • 0
    $$ \frac{a_{n+1}}{a_n} = \frac{3}{2n+3} \leq \frac{3}{2 \cdot 1 + 3} = \frac{3}{5} < 1, $$ so $a_{n+1} \leq \frac{3}{5} a_n$. Then using induction show that $a_n \leq a_1 (3/5)^{n-1} \to 0$. (Does this help?) // Alternately, in the product $$ a_n = \frac{3}{3} \cdot \frac{3}{5} \cdot \frac{3}{7} \cdots \frac{3}{2n+1}, $$ upper bound the second term onwards by $3/5$ to get $a_n \leq (\frac{3}{5})^{n-1}.$2011-10-29
  • 0
    Using the ratio test i found that $\frac{a_{n+1}}{a_n} = \frac{-3}{2n+3}$2011-10-29
  • 0
    The ratio test asks you to find the limit of that.2011-10-29
  • 0
    Added to my post a possible solution(?).2011-10-29
  • 1
    Let me suggest you post this addition to your post as an answer.2011-10-29
  • 0
    @akimo_uki: That's basically right, it is best if you use absolute values in the ratio test(it works the way you used too, but with absolute value works in more general settings). Also, the outcome is a little stronger, namely absolutely convergent.2011-10-29
  • 1
    Your "update" looks like a proper solution to me. I agree with Didier, post that as an answer. (It is perfectly acceptable to answer your own questions here.)2011-10-29
  • 0
    @user9176 the teacher didn't talk about that yet. It is in the next lecture so i didn't want to use it. Can someone post the answer? It don't have the reputation needed to do that.2011-10-29
  • 1
    You don't need any points to post an answer.2011-10-30

2 Answers 2

0

Answering my own question with the help provided.

Using the ratio test we have:

$$a_{n} = (-1)^{n+1}\frac{3^{n}}{3\cdot5\cdot7\cdots(2n+1)} $$

$$a_{n+1} = (-1)^{n+2}\frac{3^{n+1}}{3\cdot5\cdot7\cdots(2n+1)\cdot(2n+3)} $$

$$\frac{a_{n+1}}{a_{n}}=\frac{-3}{2n+3} $$

$$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_{n}}=\lim_{n\rightarrow\infty}\frac{-3}{2n+3}=\frac{3}{\infty}=0 < 1$$ Which means that the series converge.

1

As $0 and $a_{n+1} then $a_n$ converges to some point $c$. From the equality $a_{n+1}=\frac{3a_n}{2n+1}$, we have the following when $n\to\infty$:

$c=0$. So we get that $a_n\to 0$.