How to bound $\sum_{ j=1}^n j^{n-j}?$

Asked Dec 11 '21 at 19:07

Active Dec 11 '21 at 19:39

Viewed 36 times

I came across this sum in some research I'm doing. I need to bound

$$S_n=\sum_{j=1}^{n}j^{n-j}.$$

One bound I've managed is obtained by doing the following: Divide by n! and observe

$S_n/n!=\sum_{j=1}^{n}\frac{j^{n-j}}{(n)...(j+1)}\frac{1}{(j)!}\leq\sum_{j=1}^{n}\frac{1}{(j)!}\leq e-1$

So that

$S_n\leq (e-1)n!$,

but I was wondering if this can be improved at all as this seems like a pretty crude bound.

edited Dec 11 '21 at 19:39

asked Dec 11 '21 at 19:07

keagan_callis

Are you suggesting that $$\frac{j^{n-j}}{n(n-1)(\cdots (n-j+1)}\leq 1?$$ Is that obvious? Or even true? Doesn’t seem true when $j=2$ and $n$ large. – Thomas Andrews Dec 11 '21 at 19:25
Oh, I made a mistake. This is only trivially true for j < (n+1)/2. Let me redo the proof. – keagan_callis Dec 11 '21 at 19:29
I fixed it. Just made a mistake in the first equality, my bad. – keagan_callis Dec 11 '21 at 19:40
See also https://oeis.org/A026898 for more info (I found the duplicate on MSE via the OIES page). – Mike Earnest Dec 11 '21 at 19:52
Awesome! Thank you, I wasn't able to find it myself. – keagan_callis Dec 11 '21 at 19:55

0 Answers0