MHB Series Divergence: Prove $\frac{a^n}{n^23^n}$ Diverges

[Fermat1]

Prove that the series of $\frac{a^n}{n^23^n}$ diverges where $a$ is a complex number with $|a|{\geq}3$

 

[Deveno]

Please show some effort, Fermat. We have no idea what part of the problem is giving you trouble.

 

[Fermat1]

Please show some effort, Fermat. We have no idea what part of the problem is giving you trouble.

I thought about showing the sequence of terms does not tend to 0. Can you help with that?

 

[Prove It]

I thought about showing the sequence of terms does not tend to 0. Can you help with that?

I think that's a good idea. Notice that $\displaystyle \begin{align*} \frac{a^n}{n^2\,3^n} = \left( \frac{1}{n^2} \right) \left( \frac{a}{3} \right) ^n \end{align*}$. Notice that $\displaystyle \begin{align*} \frac{1}{n^2} \to 0 \end{align*}$ as $\displaystyle \begin{align*} n \to \infty \end{align*}$.

If $\displaystyle \begin{align*} \left| a \right| < 3 \end{align*}$ then $\displaystyle \begin{align*} \left| \frac{a}{3} \right| < 1 \end{align*}$ and $\displaystyle \begin{align*} \left( \frac{a}{3} \right) ^n \to 0 \end{align*}$ as $\displaystyle \begin{align*} n \to \infty \end{align*}$. Thus $\displaystyle \begin{align*} \left( \frac{1}{n^2} \right) \left( \frac{a}{3} \right) ^n \to 0 \cdot 0 = 0 \end{align*}$ as $\displaystyle \begin{align*} n \to \infty \end{align*}$ and so may converge.

If $\displaystyle \begin{align*} a = 3 \end{align*}$ then $\displaystyle \begin{align*} \frac{a}{3} = 1 \end{align*}$ and so $\displaystyle \begin{align*} \left( \frac{a}{3} \right) ^n = 1^n \to 1 \end{align*}$ as $\displaystyle \begin{align*} n \to \infty \end{align*}$, and thus $\displaystyle \begin{align*} \left( \frac{1}{n^2} \right) \left( \frac{a}{3} \right) ^n \to 0 \cdot 1 = 0 \end{align*}$ as $\displaystyle \begin{align*} n \to \infty \end{align*}$ and so may converge.

If $\displaystyle \begin{align*} a = -3 \end{align*}$ then $\displaystyle \begin{align*} \frac{a}{3} = -1 \end{align*}$ and so will oscillate between -1 and 1. Since the product of a bounded function (such as this one) and a function that goes to 0 has a limiting value of 0, that means $\displaystyle \begin{align*} \left( \frac{1}{n^2}\right) \left( \frac{a}{3} \right) ^n \to 0 \end{align*}$ as $\displaystyle \begin{align*} n \to \infty \end{align*}$ and so may converge.

Now finally, if $\displaystyle \begin{align*} \left| a \right| > 3 \end{align*}$, then $\displaystyle \begin{align*} \left| \frac{a}{3} \right| > 1 \end{align*}$ and thus $\displaystyle \begin{align*} \left( \frac{a}{3} \right) ^n \to \pm \infty \end{align*}$ as $\displaystyle \begin{align*} n \to \infty \end{align*}$ (depending on the sign of $\displaystyle \begin{align*} a \end{align*}$).

That means $\displaystyle \begin{align*} \left( \frac{1}{n^2} \right) \left( \frac{a}{3} \right) ^n \end{align*}$ gives an indeterminate form $\displaystyle \begin{align*} 0 \times \infty \end{align*}$ and so has to be transformed.

Rewrite it as $\displaystyle \begin{align*} \left( \frac{1}{n^2} \right) \left( \frac{a}{3} \right) ^n = \frac{\left( \frac{a}{3} \right) ^n}{n^2} \end{align*}$ which is now an indeterminate form $\displaystyle \begin{align*} \frac{\infty}{\infty} \end{align*}$ and so L'Hospital's Rule can be used.

$\displaystyle \begin{align*} \lim_{n \to \infty} \frac{\left( \frac{a}{3} \right) ^n}{n^2} &= \lim_{n \to \infty} \frac{\left( \frac{a}{3} \right) ^n \ln{ \left( \frac{a}{3} \right) }}{2n} \textrm{ by L'Hospital's Rule} \\ &= \lim_{n \to \infty} \frac{\left( \frac{a}{3} \right) ^n \left[ \ln{ \left( \frac{a}{3} \right) } \right] ^2 }{2} \textrm{ by L'Hospital's Rule again} \\ &\to \infty \textrm{ as } n \to \infty \end{align*}$

Since this limit is not 0 when $\displaystyle \begin{align*} |a| > 3 \end{align*}$, the series can not possibly converge when $\displaystyle \begin{align*} |a| > 3 \end{align*}$. Thus the series is divergent if $\displaystyle \begin{align*} |a| > 3 \end{align*}$.

 

[Euge]

Prove that the series of $\frac{a^n}{n^23^n}$ diverges where $a$ is a complex number with $|a|{\geq}3$

The series $\sum_{n = 1}^\infty \frac{a^n}{n^2 3^n}$ actually converges when $|a| = 3$, but diverges when $|a| > 3$.

If $|a| = 3$, then $a = 3e^{i\theta}$ for some real number $\theta$. The series reduces to

(*) $\displaystyle \sum_{n = 1}^\infty \frac{e^{in\theta}}{n^2}$.

Since $|e^{in\theta}/n^2| = 1/n^2$ for all $n$ and $\sum_{n = 1}^\infty 1/n^2$ converges (by the integral test), the series (*) converges absolutely; in particular, it converges.

If $|a| > 3$, then

$\lim_{n\to\infty} \left|\frac{a^n}{n^2 3^n}\right|^{1/n} = \lim_{n\to\infty} \frac{1}{n^{2/n}} \frac{|a|}{3} = \frac{|a|}{3} > 1$,

and so $\sum_{n = 1}^\infty \frac{a^n}{n^2 3^n}$ diverges by the root test.