# Evaluating the limit of a sequence using Squeeze Theorem

Let $a \in \mathbb R$, $0 < a < 1$. Find $$\lim_{n\to\infty}\left(\frac{a^n+a^{2n}}{1+a^3}\right)^\frac1n$$

I am supposed to use the Squeeze Theorem, so I tried the following, but I don't know how to proceed any further.

We have $\left(\frac{2a^{2n}}{1+a^3}\right)^\frac1n\le\left(\frac{a^n+a^{2n}}{1+a^3}\right)^\frac1n\le\left(\frac{2a^n}{1+a^3}\right)^{\frac1n}\implies\frac{\sqrt[n]{2}a^{2}}{(1+a^3)^\frac1n}\le\left(\frac{a^n+a^{2n}}{1+a^3}\right)^\frac1n\le\frac{\sqrt[n]{2}a}{(1+a^3)^\frac1n}$
Now, $\lim_{n\to\infty}\frac{2a^{2}}{(1+a^3)^\frac1n} = \sqrt[n]{2}a^2$ and $\lim_{n\to\infty}\frac{2a}{(1+a^3)^\frac1n} = \sqrt[n]{2}a$.
But we need both limits to be equal for the Squeeze Theorem to work.
Anyone has any ideas on how I could make some adjustments to get both limits to be equal?

• It's $\sqrt[n]{2}$, not just $2$. – user26486 Feb 22 '15 at 7:42

Since $0 < a < 1$, $a^{2n} < a^n$ for all $n$, thus

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

On the other hand

$$\left(\frac{a^n + a^{2n}}{1 + a^3}\right)^{1/n} > \left(\frac{a^n}{1 + a^3}\right)^{1/n} = a\left(\frac{1}{1 + a^3}\right)^{1/n}.$$

Since both $[2/(1 + a^3)]^{1/n}$ and $[1/(1 + a^3)]^{1/n}$ tend to $1$ as $n\to \infty$, it follows from the squeeze theorem that

$$\lim_{n\to \infty} \left(\frac{a^n + a^{2n}}{1 + a^3}\right)^{1/n} = a.$$

• Thanks! I tried this approach earlier but I made a careless mistake by using 2 instead of $\sqrt[n]{2}$. – Vizuna Feb 22 '15 at 7:50