I want to prove that if $a_n = q^n$ and $|q| < 1$ and $n \in \mathbb N$ then $\lim_{x\to \infty}a_n = 0$. My math textbook shows proof using Bernoulli inequality. I understand it, but it is quite complex. Is there any more elementary proof than this one? My idea would be to use the definition of limit and find $n$ in terms of $\epsilon$. So $|q^n - 0| < \epsilon$. This gives me $n > \frac{\log_n\epsilon}{\log_n|q|}$. Is this proof correct? If no, where is the mistake?
Proof of $\lim\limits_{x\to \infty}$
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limits
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0Did your textbook define the logarithm function before or after it proved $\lim_{n \to \infty}q^n = 0$ for $|q| < 1$? – 2017-01-21
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0It was defined before. – 2017-01-21
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0@user4201961 maybe this is the case the proof is complicated. – 2017-01-21
3 Answers
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If $q=0,$ the result is trivial. Let $\left | q\right |\neq 0$ and $\epsilon>0,$ then: $$\left | q^n-0 \right |<\epsilon\Leftrightarrow \left | q\right |^n<\epsilon \Leftrightarrow n\log \left | q\right | <\log \epsilon.\quad (*)$$ As $0<\left | q\right |<1,$ $\log \left | q\right |<0$ so the last inequality of $(*)$ is equivalent to $n>\dfrac{\log \epsilon}{\log \left | q\right |}.$ As a consequence, $\lim q^n=0.$
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Use natural logarithm: $$|q|^n<\varepsilon\iff n\ln |q|<\ln\varepsilon\iff n>\frac{\ln \varepsilon}{\ln |q|}.$$ Of course, your fraction is exactly the same but it is better to see that it really does not depend on $n$. Your proof is correct. I checked in one of calculus handbooks I have on my shelf. The proof given there is the same as yours.
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Your proof is correct. But I find it less elementary than the one that is based in Bernoulli's equality.
Nevertheless, a remark about your proof should be made, just in case. The bound for $n$ that you have found $$n>\frac{\log_n\epsilon}{\log_n|q|}$$ must not depend on $n$, and it doesn't, because $$\frac{\log_n\epsilon}{\log_n|q|}=\log_{|q|}\epsilon$$ but it is always better to make it explicit.