How to calculate $\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$

Question

I came across this strange limit whilst showing convergence of a series: $$\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$$ How can I calculate this limit?

Answer 1

HINT We have $$\dfrac12 \cdot \dfrac{3^n}{4^n} < \dfrac{2^n+3^n}{3^n+4^n} < \dfrac{3^n}{4^n}$$

Answer 2

Use the sandwich/squeeze theorem. Since

$$\frac{3^n}{2\cdot 4^n}<\frac{2^n + 3^n}{3^n + 4^n} < \frac{2\cdot 3^n}{4^n}$$

for all $n$,

$$\frac{1}{2^{1/n}} \cdot \frac{3}{4} < \sqrt[n]{\frac{2^n + 3^n}{3^n + 4^n}} < 2^{1/n}\cdot \frac{3}{4}$$

for all $n$. Since $2^{1/n} \to 1$ as $n\to \infty$, by the squeeze theorem, your limit is $3/4$.

Answer 3

The limit evaluates to... $$3 \over 4$$

Partial Proof:

$$\left({{2^n+3^n} \over {3^n+4^n}}\right)^{1/n}$$

divide the inside by $3^n$

$$\left({{(2/3)^n+1} \over {1+(4/3)^n}}\right)^{1/n}$$

$${({(2/3)^n+1)^{1/n}} \over {(1+(4/3)^n})^{1/n}}$$

The numerator evaluates to 1 and the denominator evaluates to ${4 \over 3}$ which equals ${3 \over 4}$. You'll want to verify this for yourself.

Answer 4

Squeeze theorem gives you the proof that the limit is $\frac{3}{4}$. Since you mentioned you were looking for another way to verify that the limit is correct, here is one way (although not rigorous like the squeeze theorem) $$\begin{align}\frac{2^n+3^n}{3^n+4^n} = \frac{2^n}{3^n+4^n}+\frac{3^n}{3^n+4^n} \\ = \left(\frac{\frac{1}{2^n}}{\frac{1}{2^n}}\right)\frac{2^n}{3^n+4^n}+\left(\frac{\frac{1}{3^n}}{\frac{1}{3^n}}\right)\frac{3^n}{3^n+4^n} \\ = \frac{1}{\left(\frac{3}{2}\right)^n+2^n}+\frac{1}{1+\left(\frac{4}{3}\right)^n}\end{align}$$ You should be able to see that $\lim_{n \to \infty} \left(\frac{3}{2}\right)^n+2^n = \infty$ so $\lim_{n \to \infty} \frac{1}{\left(\frac{3}{2}\right)^n+2^n} = 0$. Then notice that for large $n$ the quantity $$\frac{1}{1+\left(\frac{4}{3}\right)^n} \approx \frac{1}{\left(\frac{4}{3}\right)^n} = \left(\frac{3}{4}\right)^n$$ so for large values of $n$, $$\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}} \approx \sqrt[n]{\left(\frac{3}{4}\right)^n} = \frac{3}{4}$$

Answer 5

Hint: $$ \sqrt[\large n]{\frac{2^n+3^n}{3^n+4^n}}=\frac34\ \sqrt[\large n]{\frac{1+\left(\frac23\right)^n}{1+\left(\frac34\right)^n}} $$