Solutions - 0140-02

(a)

(1) Set up squeeze relations:

$$0\le \frac{{\cos }^{2}n}{2^n}\le \frac{1}{2^n}$$

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(2) Apply theorem:

We have:

$$\underset{n\to \infty }{\lim }\frac{1}{2^n}=0$$

Therefore:

$$\underset{n\to \infty }{\lim }\frac{{\cos }^{2}n}{2^n}=0$$

We conclude that $a_n$ converges.

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(b)

(1) Generate squeeze inequalities:

Observe:

$$3^n\le 2^n+3^n\le 3^n+3^n$$

Rewrite RHS:

$$3^n+3^n\gg \gg 2{(3)}^n$$

Raise all terms to $1/n$:

$$3\le {(2^n+3^n)}^{1/n}\le (2^{1/n})3$$

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(2) Apply squeeze theorem:

$$2^{1/n}\to 1\text{as}n\to \infty$$

Therefore:

$$(2^{1/n})3\to 3\text{as}n\to \infty$$

Conclude that: ${(2^n+3^n)}^{1/n}\to 3$