Studying the convergence of a series applying ratio's test

Question

The exercise asks me to study the convergence of the following series $$ \sum \frac{(3n)!}{3^{3n}(n!)^3}.$$

• The exercise gives me an indication to calculate $\frac{a_{n+1}}{a_n}$. So: $$\frac{a_{n+1}}{a_n}=\frac{\frac{(3(n+1))!}{3^{3(n+1)}((n+1)!)^3}}{\frac{(3n)!}{3^{3n}(n!)^3}}=\frac{(3(n+1))!3^{3n}(n!)^3}{(3n)!3^{3(n+1)}((n+1)!)^3}.$$ Let's simplify: $$\begin{align}\frac{(3(n+1))!}{(3n)!}&=\frac{(3n+3)(3n+2)(3n+1)3n(3n-1)...2·1)}{3n(3n-1)...2·1}\\&=(3n+3)(3n+2)(3n+1)\end{align}$$ $$\frac{3^{3n}}{3^{3(n+1)}}=3^{-3}$$ $$\frac{(n!)^3}{((n+1)!)^3}=\left(\frac{n!}{(n+1)!}\right)^3=\frac{1}{(n+1)^3}$$ Therefore, $\frac{a_{n+1}}{a_n}=\frac{(3n+3)(3n+2)(3n+1)}{3^3(n+1)^3}=\frac{(3n+2)(3n+1)}{3^2(n+1)^2}=\frac{9n^2+9n+2}{9n^2+18n+9}$

• Applying d'Alembert's ratio test: $$\lim \frac{a_{n+1}}{a_n}=\lim \frac{9n^2+9n+2}{9n^2+18n+9}=1$$ Then the test is inconclusive.

• Applying Raabe's test we have: $$\lim \hspace{5 pt }n \left( 1- \frac{a_{n+1}}{a_n}\right)=\lim \hspace{5 pt }n \left( 1- \frac{9n^2+9n+2}{9n^2+18n+9}\right)=\lim \hspace{5 pt }\frac{9n^2+7n}{9n^2+18n+9}=1$$ Then the test is inconclusive.

Is there an error in my reasoning or is convergence studied in another way? I am confused that I am asked to study the quotient but ratio's test and Raabe's test are inconlcusive.

Answer 1

From the OP's calculations $$\frac{a_{n+1}}{a_n}=\frac{9n^2+9n+2}{9n^2+18n+9}=1-\frac{9n+7}{9n^2+18n+9}\geq 1-\frac1n,\qquad n\geq1$$ It follows that the series diverges. Here is one useful version of Raabe's:

Raabe: Suppose$|a_n|>0$ for all $n\geq N$.

• If $\frac{|a_{n+1}|}{|a_n|}\leq 1-\frac{p}{n}$ for some $p>1$, then the series $\sum_na_n$ converges absolutely.
• If $\frac{|a_{n+1}|}{|a_n|}\geq 1-\frac{p}{n}$ for some $p\leq1$, then the series $\sum_n|a_n|$ diverges.

In many instances, one has that $$q=\lim_n n\big(1-\tfrac{|a_{n+1}|}{|a_n|}\big)$$ exists. If $q>1$, $\sum_n|a_n|$ converges while if $q<1$, $\sum_n|a_n|$ diverges. When $q=1$, the test is inconclusive.

Taking the limit to $n\rightarrow\infty$ does not help much here because $\lim_nn\Big(1-\frac{a_{n+1}}{a_n}\Big)=1$,

One can also try for the next test I the hierarchy, namely Bertran's test. $$\left(n\Big(1-\frac{a_{n+1}}{a_n}\Big)-1\right)\log n= \left(\frac{9n^2+7n}{9n^2+18n+9}-1\right)\log n=-\frac{7n+9}{9n^2+18n+9}\log n\xrightarrow{n\rightarrow\infty}0<1$$ Bertrand's test (as we have seen already) shows that $\sum_n|a_n|$ diverges.

Here is one version of Bertrand's test that is useful:

Bertrand: Suppose $|a_n|>0$ for all $n\geq N$

• (c) If $\frac{|a_{n+1}|}{|a_n|}\leq 1-\frac1n-\frac{q_n}{n\log n}$ such that $q_n\geq q>1$ for all $n\geq N$, then $\sum_na_n$ converges absolutely.
• (d) If $\frac{|a_{n+1}|}{|a_n|}\geq 1-\frac1n-\frac{q_n}{n\log n}$ such that $q_n\leq q<1$ for all $n\geq N$, then $\sum_n|a_n|$ diverges.

In many instances, $$\lim_n n\log n\big(1 -\tfrac{|a_{n+1}|}{|a_n|}\big)-\log n =q $$ exists. If $q>1$, then $\sum_n|a_n|$ converges while if $q<1$, $\sum_n|a_n|$ diverges. When $q=1$, the test is inconclusive.

Answer 2

By your own calculations, Raabe's test is conclusive: $$n\left(\frac{a_n}{a_{n+1}}-1\right)=n\left(\frac{9n^2+18n+9}{9n^2+9n+2}-1\right)=\frac{9n^2+7n}{9n^2+9n+2}<1$$ hence the series is divergent.