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Find the $\limsup (n+2^n)^{1/n}$.

I found that this is $2$ by calculating the limit of the ratio of $a_n$ and $a_{n+1}$. But I was wondering if there is a direct way to find the limit without considering the ratios. I would greatly appreciate any help.

2 Answers 2

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It is the reciprocal of the radius of convergence of the following series by the root test:

$$\sum_{n=0}^\infty(n+2^n)x^n$$

Which clearly has radius of convergence $1/2$. Thus, the correct $\limsup$ is $2$.

Another approach is squeeze:

$$(2^n)^{1/n}<(n+2^n)^{1/n}<(2^n+2^n)^{1/n}$$

Which tells us the limit exists, which is equal to the limit superior.

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    I initially got this question from finding the radius of convergence of the above series. But how is it clear that the radius is $1/2$? – 2017-02-28
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    @takecare It is a geometric series plus an almost geometric series. You could also instead consider my squeezed limit, which might prove to be much more intuitive. – 2017-02-28
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Forget about the $\limsup$. Let's find the limit. Since $n\le 2^n$ for $n\in \mathbb N,$ we have

$$2 = (2^n)^{1/n} \le (n+2^n)^{1/n} \le (2^n+2^n)^{1/n} = (2^{n+1})^{1/n} = 2^{(n+1)/n}.$$

The right side of this $\to 2^1 = 2$ and we see our limit is $2.$