- #1
evinda
Gold Member
MHB
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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 $a<-1$ ?
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)
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)