MHB Does the sequence $(a^n b^{n^2})$ converge for all values of $a$ and $b$?

[evinda]

Hello! (Wave)

I want to check as for the convergence the sequence $(a^n b^{n^2})$ for all the possible values that $a,b$ take.

I have thought the following:

We have that $\lim_{n \to +\infty} a^n=+\infty$ if $a>1$, $\lim_{n \to +\infty} a^n=0$ if $-1<a<1$, right?

What happens for $a<-1$ ?

We have that $a^n b^{n^2}=(ab^n)^n$.

If $a>0$, $b>1$ then $\lim_{n \to +\infty} (ab^n)=+\infty$ and thus $\lim_{n \to +\infty} (ab^n)^n=+\infty$.

If $a<0$, $b>1$, then $\lim_{n \to +\infty} (ab^n)=-\infty$. Then what can we say about $\lim_{n \to +\infty} (ab^n)^n$ ?

If $0<b<1$, then $\lim_{n \to +\infty} (ab^n)=0$ and thus $\lim_{n \to +\infty} (ab^n)^n=\lim_{n \to +\infty} 0=0$.

Am I right so far?

If $b<0$, what can we say about the desired limit? (Thinking)

 

[GJA]

Hi evinda,

Your analysis is certainly on the right track. By using two facts we can reduce the number of cases that need to be considered:

1. $\displaystyle\lim_{n}|x_{n}|=0$ if and only if $\displaystyle\lim_{n} x_{n}=0.$
2. If $\displaystyle\lim_{n}|x_{n}|=\infty$, then $x_{n}$ is an unbounded sequence.

The first statement can be proved using the definition of the limit of a sequence. The second can be established via contradiction and the definition of a bounded sequence.

For $a=0$ or $b=0$ we see that the sequence converges automatically. Thus, in what follows we always consider $a\neq 0$ and $b\neq 0$.

Case 1: $|b|<1.$

By writing $y_{n}=(ab^{n})^{n}$ we see that $$|y_{n}|=e^{n\ln|a|+n^{2}\ln|b|}.$$ Since $|b|<1$, $\ln|b|<0$ and the presence of the term $n^{2}$ above will guarantee the dominance of $n^{2}\ln|b|.$ Hence, $$\lim_{n}|y_{n}|= e^{-\infty}= 0.$$ Using Fact 1 above, we have $\displaystyle\lim_{n}(ab^{n})^{n}=0.$

Case 2: $|b|>1.$

We have $|ab^{n}|=|a||b|^{n}\rightarrow\infty~~\Longrightarrow~~|(ab^{n})^{n}|\rightarrow\infty.$ Using Fact 2, it follows that $(ab^{n})^{n}$ is unbounded and, therefore, must necessarily diverge.

Remaining Cases: $b=\pm 1.$

I will leave this to you.

Hopefully by reducing the number of cases the problem is more tractable. Let me know if anything requires further clarification.

 

[evinda]

Hi evinda,

Your analysis is certainly on the right track. By using two facts we can reduce the number of cases that need to be considered:

1. $\displaystyle\lim_{n}|x_{n}|=0$ if and only if $\displaystyle\lim_{n} x_{n}=0.$
2. If $\displaystyle\lim_{n}|x_{n}|=\infty$, then $x_{n}$ is an unbounded sequence.

The first statement can be proved using the definition of the limit of a sequence. The second can be established via contradiction and the definition of a bounded sequence.

For $a=0$ or $b=0$ we see that the sequence converges automatically. Thus, in what follows we always consider $a\neq 0$ and $b\neq 0$.

Case 1: $|b|<1.$

By writing $y_{n}=(ab^{n})^{n}$ we see that $$|y_{n}|=e^{n\ln|a|+n^{2}\ln|b|}.$$ Since $|b|<1$, $\ln|b|<0$ and the presence of the term $n^{2}$ above will guarantee the dominance of $n^{2}\ln|b|.$ Hence, $$\lim_{n}|y_{n}|= e^{-\infty}= 0.$$ Using Fact 1 above, we have $\displaystyle\lim_{n}(ab^{n})^{n}=0.$

Case 2: $|b|>1.$

We have $|ab^{n}|=|a||b|^{n}\rightarrow\infty~~\Longrightarrow~~|(ab^{n})^{n}|\rightarrow\infty.$ Using Fact 2, it follows that $(ab^{n})^{n}$ is unbounded and, therefore, must necessarily diverge.

I see... (Yes)

Remaining Cases: $b=\pm 1.$

I will leave this to you.

Hopefully by reducing the number of cases the problem is more tractable. Let me know if anything requires further clarification.

For $b=1$, the convergence of the sequence $(ab^n)^n$ equals to the convergence of $a^n$ and for $b=-1$ the convergence of the sequence $(ab^n)^n$ equals to the convergence of $a^n(-1)^{n^2}$.

It doesn't hold that $a^n(-1)^{n^2}$ converges iff $a^n$ converges, does it? (Thinking)

 

[GJA]

For $b=1$, the convergence of the sequence $(ab^n)^n$ equals to the convergence of $a^n$ and for $b=-1$ the convergence of the sequence $(ab^n)^n$ equals to the convergence of $a^n(-1)^{n^2}$.

This is correct.

It doesn't hold that $a^n(-1)^{n^2}$ converges iff $a^n$ converges, does it? (Thinking)

This is not true - take $a=1$ as a counterexample.

 

[evinda]

This is correct.

Nice! (Smirk)

This is not true - take $a=1$ as a counterexample.

Ok, how do we show that $(-1)^{n^2}$ does not converge? (Thinking)

 

[GJA]

Ok, how do we show that $(-1)^{n^2}$ does not converge? (Thinking)

This is the alternating sequence $\{-1, 1, -1, 1,\ldots\}$ because $n^{2}$ is even iff $n$ is even. Moreover, the sequence possesses two convergent subsequences: $\{-1, -1, -1, \ldots\}$ and $\{1, 1, 1, \ldots\}$, which cannot happen for a convergent sequence.

 

[evinda]

This is the alternating sequence $\{-1, 1, -1, 1,\ldots\}$ because $n^{2}$ is even iff $n$ is even. Moreover, the sequence possesses two convergent subsequences: $\{-1, -1, -1, \ldots\}$ and $\{1, 1, 1, \ldots\}$, which cannot happen for a convergent sequence.

I see... Thanks a lot! (Happy)