Check the convergence of the series $\sum_{n=1}^\infty \frac{(3n-2)!!!}{3^n n!}$ and $\sum_{n=1}^\infty (-1)^n\frac{(3n-2)!!!}{3^n n!}$?

Question

Check the convergence of the following series $$\sum_{n=1}^\infty \frac{(3n-2)!!!}{3^n n!}$$ and $$\sum_{n=1}^\infty (-1)^n\frac{(3n-2)!!!}{3^n n!}$$

My attempt:

I tried Ratio test. I got \begin{align} \lim_{n\to \infty} \left| \frac{a_{n+1}}{a_{n}} \right| &= \lim_{n\to \infty} \left| \frac{\dfrac{(3n+1)!!!}{3^{n+1} (n+1)!}}{\dfrac{(3n-2)!!!}{3^n n!}} \right| \\ &= \lim_{n\to \infty} \left| \frac{\dfrac{(3n-2)!!!(3n+1)}{3^{n}\cdot 3 (n)!(n+1)}}{\dfrac{(3n-2)!!!}{3^n n!}} \right| \\ &= \lim_{n\to \infty} \left| \frac{\dfrac{(3n+1)}{3 (n+1)}}1 \right| \\ &=1. \end{align}

The test is inconclusive.

When I took Generalized Raabe’s test,

I got $$\lim_{n\to \infty}n\left(\left|\frac{3 (n+1)}{(3n+1)}\right|-1\right)=2/3.$$

This shows that series is not absolutely convergent.

From the normal Raabe’s test, it is clear that $\sum_{n=1}^\infty \frac{(3n-2)!!!}{3^n n!}$ is divergent. I don’t know how do I conclude for the case $\sum_{n=1}^\infty (-1)^n\frac{(3n-2)!!!}{3^n n!}$. Please help me.

Answer 1

Note that $$ \sum_{n=1}^\infty \frac{(3n-2)!!! x^n}{3^n n!} = -1+\sum_{n=0}^\infty \binom{-1/3}{n} (-1)^n x^n =-1+(1-x)^{-1/3}\qquad\text{for } |x|<1 . $$ When $x \to 1^-$, conclude that the series at $x=1$ diverges to $+\infty$. When $x = -1$, the series converges to $-1+2^{-1/3}$ by Abel's convergence theorem.

Answer 2

Define a sequence $\{a_n\}_{n\in\mathbb N}$ as follows. $$a_{n} =\frac{( 3n-2) !!!}{3^{n} \cdotp n!}$$

As you did in your ratio test, $$\begin{aligned} a_{n+1} & =\frac{( 3n+1) !!!}{3^{n+1} \cdotp ( n+1) !}\\ & =\frac{3n+1}{3( n+1)} \cdotp \frac{( 3n-2) !!!}{3^{n} \cdotp n!}\\ & =\left( 1-\frac{2}{3n+3}\right) \cdotp a_{n} \end{aligned}$$

It follows that: $$\frac{a_{n+1}}{a_{n}} < 1\Longrightarrow a_{n+1} < a_{n}$$

Thus, $\{a_n\}$ is monotonically decreasing. It is also bounded below by $0$ because all terms are positive.

By monotone convergence theorem, $\{a_n\}$ is convergent.

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Now show that $\lim a_n=0$.

$$a_{n} =\frac{1}{3} \cdot \frac{4}{6} \cdot \frac{7}{9} \cdots \frac{3n-2}{3n}$$

Consider the sequence $$b_n=\left(\frac{1}{n}\left(\frac{1}{3} +\frac{4}{6} +\frac{7}{9} +\cdots +\frac{3n-2}{3n}\right)\right)^{n}$$

$a_1=b_1$, however, for all $n\geq 2$, $0<a_n<b_n\tag*{}$ (by A.M.-G.M. inequality).

Show that $\lim b_n=0$ and conclude by sandwich theorem that $\lim a_n=0$.

$$\begin{aligned} b_{n} & =\left(\frac{1}{n}\sum _{j=1}^{n}\left( 1-\frac{2}{3j}\right)\right)^{n}\\ & =\left( 1-\frac{2}{3n} \cdot \sum _{j=1}^{n}\frac{1}{j}\right)^{n} \end{aligned}$$

$$\Rightarrow \lim b_n=\exp\left(-\frac{2}{3}\lim \sum _{j=1}^{n}\frac{1}{j}\right)$$

This is a partial sum sequence of a well-known divergent series.

$$\therefore \lim b_n=\lim_{x\to\infty}\exp\left(-\frac{2}{3}x\right)=0$$

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Conclusions:

• $\{a_n\}$ is a seq. of positive reals.
• It is monotonically decreasing.
• It converges to $0$.

These conditions are sufficient for $\sum(-1)^n a_n$ to converge according to Leibniz's test. $\blacksquare$

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I found a simpler way to find $\lim a_n$ here; thanks to user Riemann.

Let $$a_{n}=\frac{1}{3}\cdot\frac{4}{6}\cdot\frac{7}{9}\cdots\frac{3n-2}{3n}$$ $$b_{n}=\frac{2}{4}\cdot\frac{5}{7}\cdot\frac{8}{10}\cdots\frac{3n-1}{3n+1}$$ $$c_{n}=\frac{3}{5}\cdot\frac{6}{8}\cdot\frac{9}{11}\cdots\frac{3n}{3n+2}$$ obviously, $$a_n<b_n<c_n,\quad \forall n=1,2,\cdots,$$ so $$\begin{align}0<a_n &=\sqrt[3]{(a_n)^3}\\ &<\sqrt[3]{a_nb_nc_n}\\ &=\underbrace{\sqrt[3]{\frac{1}{3n+1}\cdot\frac{2}{3n+2}}}_{\lim\ =\ 0}\end{align}$$

This sequence converges to $0$ so again, you can conclude $\lim a_n=0$ by the sandwich theorem.