convergence of $(2^n + 3^n)^{\frac{1}{n}}$

Question

Prove that the sequence $\{(2^n + 3^n)^{\frac{1}{n}}\}$ converges and find its limit.

I've deduced that the sequence will converge to $3$, just need help writing a formal proof of this fact.

Answer 1

Hint:

$$3 = (3^n)^{1/n} < (2^n + 3^n)^{1/n} < (2\cdot 3^n)^{1/n} = 3 \cdot 2^{1/n}$$

Now squeeze.

Answer 2

Factor out $3$ from the expression: $$ (2^n+3^n)^{\frac1n}=3\biggl(1+\Bigl(\frac23\Bigr)^n\biggr)^{\frac1n} $$ The second factor tends to $1$ as $n\to\infty$, since $$\ln\biggl(1+\Bigl(\frac23\Bigr)^n\biggr)^{\frac1n}=\frac1n\,\ln\biggl(1+\Bigl(\frac23\Bigr)^n\biggr)\sim_\infty \frac 1n\Bigl(\frac23\Bigr)^n $$ and both factors tend to $0$.

More generally you can prove this way: $$\lim_{n\to\infty}(a^n+b^n)^{\frac1n}=\max(a,b).$$

Answer 3

For an upper bound: $$\left(2^n+3^n\right)^\frac{1}{n} \leq \left(3^n+3^n\right)^\frac{1}{n} = 3\sqrt[n]{2}$$

$$\lim_{n\to\infty} 3\sqrt[n]{2} = 3$$

$$\lim_{n\to\infty} \left(2^n+3^n\right)^\frac{1}{n} \leq 3$$

For a lower bound:

$$\left(2^n+3^n\right)^\frac{1}{n} \geq \left(3^n\right)^\frac{1}{n} = 3$$

$$\lim_{n \to\infty} \left(2^n+3^n\right)^\frac{1}{n} \geq 3$$

It's both less than or equal to $3$ and greater than or equal to $3$, so it must be equal to $3$.