$\lim_{n\to \infty} {\sqrt[n]{n^{-n}+2^n}}$?

Question

$\lim_{n\to \infty} {\sqrt[n]{n^{-n}+2^n}}$

Intuitively, this seems like it should equal 2, but how would one go about showing this? I have tried factoring this somehow, but whatever form I get it in has some type of addition in the radicand. What other techniques could I use?

Do you know about continuity? All of the maps here are continuous for $n>0$. — JP McCarthy, Apr 01 '14 at 12:36
@StephenMontgomery-Smith, you mean $\lim_{n\to\infty} 2(1+(1/(2n))^n)^{1/n}$, right? Once it is expressed in this manner, then the limit is clearly 2 (you can use a trick with the natural logarithm if you want). — Nicholas Stull, Apr 01 '14 at 12:44

score 4 · Accepted Answer · answered Apr 01 '14 at 12:36

4

Hint

$$2^n \leq n^{-n}+2^n \leq 2^{n+1}$$

answered Apr 01 '14 at 12:36

N. S.

132,525

score 0 · Answer 2 · edited Apr 13 '17 at 12:21

We can use the fact that if $a_n>0$ for all $n\ge1$ and the sequence $\frac{a_{n+1}}{a_n}$ converges in $[0,\infty]$, then $$ \lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n} $$ (see this question).

We have that $$ \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\biggl(\frac{(n+1)^{-(n+1)}+2^{n+1}}{n^{-n}+2^n}\biggr)=\lim_{n\to\infty}\frac{(n+1)^{-(n+1)}}{n^{-n}+2^n}+\lim_{n\to\infty}\frac2{\frac{n^{-n}}{2^n}+1}=0+2 $$ and $$ \lim_{n\to\infty}a_n^{1/n}=2. $$

$\lim_{n\to \infty} {\sqrt[n]{n^{-n}+2^n}}$?

2 Answers2