I am trying to determine how to find the tightest upper and lower bounds when using the squeeze theorem. If there is a general technique to determining them I would appreciate the insight. Specifically I have this problem:

$$\lim_{n\to \infty} \sqrt[n]{2\left(\frac12\right)^n+\left(\frac23\right)^n+3\left(\frac12\right)^n}$$

I know that:

$$\sqrt[n]{\left(\frac23\right)^n}\le \sqrt[n]{2\left(\frac12\right)^n+\left(\frac23\right)^n+3\left(\frac12\right)^n} \le \sqrt[n]{6\left(\frac12\right)^n}$$

What steps were taken to find the upper bound?


1 Answer 1


You want to bound your expression above and below with other expressions that are simpler and converge to the same limit. Since we have a $n$-th root and terms raised to the $n$-th power, it seems reasonable to try and use the fact that $\sqrt[n]{x^n}=x$.

For the lower bound simply note that $\sqrt[n]{x}$ is increasing and $a^n\ge 0$ for $a>0$. Then, $$ \sqrt[n]{2\left(\frac12\right)^n+\left(\frac23\right)^n+3\left(\frac12\right)^n} \ge 0+\sqrt[n]{\left(\frac23\right)^n+0}=\frac23$$ For the upper bound note that $\left(\frac12\right)^n\le \left(\frac23\right)^n$ for $n>0$. Therefore, $$ \sqrt[n]{2\left(\frac12\right)^n+\left(\frac23\right)^n+3\left(\frac12\right)^n} \le \sqrt[n]{2\left(\frac23\right)^n+\left(\frac23\right)^n+3\left(\frac23\right)^n}=\frac23\sqrt[n]{6}$$ The squeeze theorem will now do the trick (note that $\sqrt[n]a\to 1$)

  • $\begingroup$ This looks much better now! $\endgroup$
    – amWhy
    Dec 6, 2012 at 17:38
  • $\begingroup$ @amWhy It sure does sir! $\endgroup$
    – Nameless
    Dec 6, 2012 at 17:40

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