I want to calculate the limit of: limn→∞(nn+n!e1/n2n+nn+1)1/n\lim_{n \to \infty}{\left(\frac{n^n+n!e^{1/n}}{2^n+n^{n+1}}\right)^{1/n}} [on hold]

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}}\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 11, but I am having trouble getting to it. Any ideas would be greatly appreciated.



4 Answers


We know e^{1/n} \to 1e^{1/n} \to 1 and n!\le n^n.n!\le n^n. So for large nn the expression is bounded above by

\left (\frac{n^n + 2n^n}{n^{n}}\right )^{1/n} = 3^{1/n} \to 1.\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}2^n < n^{n+1} for large nn 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.\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 1.1.      I'm waiting to see if OP knows how to conclude that n^{1/n} \to 1n^{1/n} \to 1. – Brian Tung Oct 20 at 19:59      Perfect! Thanks. 🙂 Though I will have to prove those inequalities, but that shouldn't be hard. – Mykybo Oct 20 at 20:11      In this equation: \left (\frac{n^n + 2n^n}{n^{n}}\right )^{1/n}\left (\frac{n^n + 2n^n}{n^{n}}\right )^{1/n}, what happened with 2^n2^n in the denominator? – Mykybo Oct 20 at 20:26      I removed it, making the denominator smaller, hence the fraction larger – zhw. Oct 20 at 20:27 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}}\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)= \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}\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!n^n > n! for all n>1n>1, we have, for the RHS of (1)(1), that
\log\left(\frac1n+\frac{n!e^{1/n}}{n^{n+1}}\right) \leq \log\left(\frac{e^{1/n}+1}{n}\right)\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 nn 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{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{ne^{-1/n}+n}{n^2+n^2e^{-1/n}} = -\lim_{n\to\infty}\frac{e^{-1/n}+1}{n(e^{-1/n}+1)}

If we go back to (1)(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\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)(1), we now look at the LHS. Dividing by nn, 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 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}=\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 00, as the denominator grows far faster than the numerator. Again going back to (1)(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\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}\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}


Use Stirling’s formula to get an equivalent of n!n!:

n!\sim \sqrt{2\pi n}\left(\frac ne\right)^n.n!\sim \sqrt{2\pi n}\left(\frac ne\right)^n.



Do we even need this? Isn’t the part in parentheses dominated by \frac{n^n}{n^{n+1}}\frac{n^n}{n^{n+1}}?
– Brian Tung
Oct 20 at 19:21



@BrianTung I’m not sure of that, since 2^n+n^{n+1}>n^{n+1}2^n+n^{n+1}>n^{n+1}…
– E. Joseph
Oct 20 at 19:24



We haven’t defined (‘learned’) Stirling’s formula yet, so I cannot use that. We have learned only the basic criterias for solving limits so far.
– Mykybo
Oct 20 at 19:27



@Mykybo alright. This can be done using inequalities and some theorems using compositions of functions… I’ll see if I have time to throw an answer together
– Brevan Ellefsen
Oct 20 at 19:35



Thanks a lot 🙂
– Mykybo
Oct 20 at 19:38

First, we divide both numerator and denominator by n^nn^n to obtain

= \left[\frac{1+\frac{n!e^{1/n}}{n^n}}{\left(\frac{2}{n}\right)^n+n}\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)1+o(1/n), and the denominator is n+o(1/n)n+o(1/n), so the fraction as a whole goes to 1/n1/n in the limit as n \to \inftyn \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

\lim_{n \to \infty} \left(\frac{1}{n}\right)^\frac{1}{n} = \lim_{x \to 0} x^x

Can you take it from there?