Show that the sequence $a_n = \frac{2^n+\cos{(n\pi)}3^n}{5^n}$ converges to $0$.
Would we use the squeeze theorem in this case? Please help.
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$\begingroup$ Did you try to use arithmetic of limits? Do you know that $a^n\to 0$ when $0<a<1$? $\endgroup$– MarkSep 27, 2019 at 23:06
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2 Answers
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This is how would I prove it: $$-1 \leq \cos (n \pi) \leq 1$$ When $\cos(n \pi)=-1$,
$$\lim_{n \to \infty}\frac{2^n-3^n}{5^n}=0$$ as $2^n-3^n \leq 5^n$, when $n \geq -1$ and both $2^n-3^n$ amd $5^n$ are motonically increasing.
When $\cos(n \pi)=1$,
$$\lim_{n \to \infty}\frac{2^n+3^n}{5^n}=0$$ as $2^n+3^n \leq 5^n$, when $n \geq 1$ and both $2^n+3^n$ amd $5^n$ are motonically increasing.
So, $$\lim_{n \to \infty}\frac{2^n+ \cos(n \pi)3^n}{5^n}=0$$ This is in no way rigorous, just my thoughts.
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Gift:
$$\cos(n\pi)=(-1)^n\quad\quad$$ then $$a_n=\left(\dfrac25\right)^n+\left(-\dfrac35\right)^n$$
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$\begingroup$ u dont need gift, just $||\cos||_\infty \le 1$ $\endgroup$ Sep 27, 2019 at 23:19
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$\begingroup$ @mathworker21 The point is that by doing this, the limit becomes more immediate without having to appeal to all the write-up that would be required of a Squeeze Theorem solution. $\endgroup$ Sep 28, 2019 at 0:06
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$\begingroup$ How would we do it using the squeeze theorem? $\endgroup$– squenshlSep 28, 2019 at 5:40