Asymptotics of a function

I hope this question is not too simple, but I would like to know the asymptotic behaviour of the following function $f: \mathbb{N}^{+} \rightarrow \mathbb{Q}$ where $$f(n) = \sum_{i=1}^{n} \frac{i^n}{n^{4i}}$$ Any references, pointers, or answers would be most appreciated.

• $f(n)=n^{n+o(n)}$.
– Did
May 8, 2013 at 12:03
• Thanks Didier - that certainly agrees with the data. Would you mind briefly sketching your reasoning? May 8, 2013 at 12:38
• Which context did you meet this beast in?
– Did
May 8, 2013 at 15:44
• It popped up in the analysis of an algorithm. According to Maple the corresponding integral can be expressed in terms of a Whittaker function. May 8, 2013 at 15:54
• You should be able to get this with the Laplace method for sums - find the dominant term, convert the Riemann sum to an integral of the form $\int dx \exp(f(x-x_0))$ and use Laplace's method to estimate the integral. See Bender and Orszag's book for an example. May 8, 2013 at 18:19

From the comments so far (including mine above) it follows that $$f(n) = \sum_{i=1}^{n} \frac{i^n}{n^{4i}}=\frac{n!}{(4\ln n)^{n+1}}\left(1+O(n^{-1/2}\ln n)\right).$$

• and by Stirling's formula this gives us $f(n)=\sqrt{2 \pi n} (\frac{n}{4e \ln n})^n (1+O(n^{-1/2}))$. May 9, 2013 at 9:49
• @Johan: Yes. In fact the error term I indicated comes from Stirling's formula. Either way, you have to compare $(n/e)^n$ with $n!$. May 9, 2013 at 10:18
• @Johan: Note that for you version we need the refined Stirling's formula with explicit error term. May 9, 2013 at 10:20
• I'm pretty sure the power of $4\ln n$ should be $n+1$ and not $n$. May 9, 2013 at 11:58
• And the error term is exponentially small. The Euler-Maclaurin formula will show it. May 9, 2013 at 12:05

$f(n) = n^n 4^{-4n} (1 + O(n^{-4}))$. The sum is strongly dominated by its last term. I hope this isn't homework.

Apologies. I misread the question exact as Benoît suggests.

• Didn't you took a $n$ for a $4$? The last term is $n^{-3n}$. May 8, 2013 at 15:21
• The last term is $n^{-3n}$, which is certainly not dominant! The asymptotics is probably obtainable by comparing with $$\int_0^\infty x^n n^{-4x}\,dx=\frac{n!}{(4\ln n)^n}.$$ May 8, 2013 at 15:21
• The main contribution is around $i=n/(4\log n)$, not around $i=n$.
– Did
May 8, 2013 at 15:48
• @Michael: I think your observation will do nicely (having bounded the contribution of the tail $x > n$). Thanks! May 8, 2013 at 15:55