I want to calculate the limit of: $\lim_{n \to \infty}{\left(\frac{n^n+n!e^{1/n}}{2^n+n^{n+1}}\right)^{1/n}}$ I want to calculate the limit of:
$$\lim_{n \to \infty}{\left(\frac{n^n+n!e^{1/n}}{2^n+n^{n+1}}\right)^{1/n}}$$
or prove that it does not exist. Now I know the result is $1$, but I am having trouble getting to it. Any ideas would be greatly appreciated.
 A: We know $e^{1/n} \to 1$ and $n!\le n^n.$ So for large $n$ the expression is bounded above by
$$\left (\frac{n^n + 2n^n}{n^{n}}\right )^{1/n} = 3^{1/n} \to 1.$$
On the other hand, because $2^n < n^{n+1}$ for large $n$ the expression is at least
$$\left (\frac{n^n}{2n^{n+1}}\right )^{1/n}= \frac{1}{2^{1/n}\cdot n^{1/n}} \to 1.$$
By the squeeze theorem the limit is $1.$
A: Here is the answer I said I would provide earlier. I try to rigorously explain in depth what I am doing in each step, and explicitly state when I leave a proof of something non-trivial out
$$\lim_{n \to \infty}{\left(\frac{n^n+n!e^{1/n}}{2^n+n^{n+1}}\right)^{1/n}}$$
$$= \exp\left(\lim_{n \to \infty}\frac 1n \log\left(\frac{n^n+n!e^{1/n}}{2^n+n^{n+1}}\right)\right)$$
If we now focus on
$$\log\left(\frac{n^n}{2^n+n^{n+1}}\right)\leq\log\left(\frac{n^n+n!e^{1/n}}{2^n+n^{n+1}}\right)\leq\log\left(\frac{n^n+n!e^{1/n}}{n^{n+1}}\right) \tag{1}$$
If we accept that $n^n > n!$ for all $n>1$, we have, for the RHS of $(1)$, that
$$\log\left(\frac1n+\frac{n!e^{1/n}}{n^{n+1}}\right) \leq \log\left(\frac{e^{1/n}+1}{n}\right)$$
If we now divide by $n$ and take the limit (using L'Hopital and noting the numerator goes to infinty...I'll let you prove this if you wish to do so!) we get:
$$\lim_{n\to\infty}\frac{1}{n}\log\left(\frac{e^{1/n}+1}{n}\right)=-\lim_{n\to\infty}\frac{ne^{-1/n}+n+1}{n^2+n^2e^{-1/n}}$$
$$=-\lim_{n\to\infty}\frac{ne^{-1/n}+n}{n^2+n^2e^{-1/n}} = -\lim_{n\to\infty}\frac{e^{-1/n}+1}{n(e^{-1/n}+1)}$$
$$=-\lim_{n\to\infty}\frac{1}{n}=0$$
If we go back to $(1)$ we get the following bound on the RHS:
$$\lim_{n\to \infty}\frac 1n\log\left(\frac{n^n+n!e^{1/n}}{2^n+n^{n+1}}\right) \leq \lim_{n\to \infty}\frac 1n\log\left(\frac{e^{1/n}+1}{n}\right)=0$$
Going back up to $(1)$, we now look at the LHS. Dividing by $n$, we can trivially apply L'Hopital's Rule, rearrange, and get the following equalities:
$$\lim_{n\to\infty} \frac 1n\log\left(\frac{n^n}{2^n+n^{n+1}}\right)=\lim_{n\to\infty}\frac{2^n (\log(n/2)+1)-n^n}{n^{n+1}+2^n}$$
$$=\lim_{n\to\infty}\frac{2^n \log(n/2)+2^n}{n^{n+1}+2^n}=\lim_{n\to\infty}\frac{\log(n/2)+1}{n\left(\frac n2\right)^n+1}$$
The last limit clearly goes to $0$, as the denominator grows far faster than the numerator. Again going back to $(1)$, we now have the lower bound
$$\lim_{n\to \infty}\frac 1n\log\left(\frac{n^n+n!e^{1/n}}{2^n+n^{n+1}}\right)>\lim_{n\to\infty} \frac 1n\log\left(\frac{n^n}{2^n+n^{n+1}}\right)=0$$
Applying the squeeze theorem, we can now deduce the original limit
$$\color{red}{\exp\left(\lim_{n \to \infty}\frac 1n \log\left(\frac{n^n+n!e^{1/n}}{2^n+n^{n+1}}\right)\right) = \exp(0) = 1}$$
A: Hint
Use Stirling's formula to get an equivalent of $n!$:
$$n!\sim \sqrt{2\pi n}\left(\frac ne\right)^n.$$
A: First, we divide both numerator and denominator by $n^n$ to obtain
$$
\left(\frac{n^n+n!e^{1/n}}{2^n+n^{n+1}}\right)^{1/n}
    = \left[\frac{1+\frac{n!e^{1/n}}{n^n}}{\left(\frac{2}{n}\right)^n+n}\right]^{1/n}
$$
Now note that the numerator is $1+o(1/n)$, and the denominator is $n+o(1/n)$, so the fraction as a whole goes to $1/n$ in the limit as $n \to \infty$.  The desired limit is thus
$$
\lim_{n \to \infty} \left(\frac{1}{n}\right)^\frac{1}{n} = \lim_{x \to 0} x^x
$$
Can you take it from there?
