# Show that the sequence $a_n = \frac{2^n+\cos{(n\pi)}3^n}{5^n}$ converges to $0$. [closed]

Show that the sequence $$a_n = \frac{2^n+\cos{(n\pi)}3^n}{5^n}$$ converges to $$0$$.

• Did you try to use arithmetic of limits? Do you know that $a^n\to 0$ when $0<a<1$?
– Mark
Sep 27, 2019 at 23:06

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.

• Thank you very much looks good to me! Sep 28, 2019 at 0:06

$$\cos(n\pi)=(-1)^n\quad\quad$$ then $$a_n=\left(\dfrac25\right)^n+\left(-\dfrac35\right)^n$$
• u dont need gift, just $||\cos||_\infty \le 1$ Sep 27, 2019 at 23:19