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

In summary, the question is about the convergence of the sequence $(a^n b^{n^2})$ for all possible values of $a$ and $b$. By using some known facts, the cases can be reduced to consider $a\neq 0$ and $b\neq 0$. For $|b|<1$, the sequence converges to 0. For $|b|>1$, the sequence is unbounded and therefore diverges. For $b=1$, the convergence is equivalent to that of $a^n$, but for $b=-1$, the convergence is not equivalent to that of $a^n$. The sequence $(-1)^{n^2}$ does not converge as
  • #1
evinda
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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 $a<-1$ ? :confused:

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)
 
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  • #2
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.
 
  • #3
GJA said:
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)

GJA said:
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)
 
  • #4
evinda said:
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.

evinda said:
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.
 
  • #5
GJA said:
This is correct.

Nice! (Smirk)

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

Ok, how do we show that $(-1)^{n^2}$ does not converge? (Thinking)
 
  • #6
evinda said:
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.
 
  • #7
GJA said:
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)
 

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

What is the definition of "convergence of sequence"?

The convergence of a sequence refers to the behavior of the terms in a sequence as the number of terms approaches infinity. In other words, it describes what happens to the terms in a sequence as the sequence gets longer and longer.

How is the convergence of a sequence determined?

The convergence of a sequence is typically determined by analyzing the behavior of the terms in the sequence, such as whether they approach a specific value or whether they become increasingly smaller. Other mathematical tools, such as limit theorems, can also be used to determine convergence.

What are the different types of convergence of a sequence?

The three main types of convergence of a sequence are pointwise convergence, uniform convergence, and absolute convergence. Pointwise convergence means that each individual term in the sequence approaches a specific value, while uniform convergence means that the entire sequence approaches a specific value. Absolute convergence is a specific type of uniform convergence that also takes into account the absolute value of the terms in the sequence.

How does the convergence of a sequence relate to its limit?

The limit of a sequence is the value that the terms in the sequence approach as the number of terms approaches infinity. If the sequence is convergent, then the limit is equal to the value that the terms approach. However, not all sequences have a limit, and some may have different limits depending on the type of convergence.

In what real-world applications is the concept of convergence of sequence used?

The concept of convergence of sequence is used in many areas of science and mathematics, such as in calculus, physics, and computer science. It is essential for understanding the behavior of systems that involve continuously changing variables, such as the motion of objects or the convergence of algorithms. It is also used in data analysis, such as in time series analysis, to identify patterns and trends in data.

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