I think I got it. Let me know if there are any errors.
The method is to prove, by induction, that |F(n)| < (1/e)^n. This will then show that F(n) --> 0 by the squeeze theorem as e^(-n) --> 0.
n=1. |F(1)|=e-8/3 ~ 0.052. (1/e)^1 ~ 0.368. Therefore, |F(1)| < (1/e)^1.
Now assume that |F(n)| < (1/e)^n.
We want to have |F(n+1)| < (1/e)^(n+1).
Let G(n)=(2n+2/3-F(n))*e^(-n); i.e., the sum part of F(n).
F(n)=2n + 2/3 - G(n) e^n.
Then by induction,
-(1/e)^n < 2n+2/3 - G(n) e^n < (1/e)^n.
I will solve this for G(n) and then let n=n+1.
Check my arithmetic but I think G(n) can be bounded by, from that last inequality, the following two extremes:
e^(-n) (2n+2/3-e^(-n)) < G(n) < e^(-n) (2n+2/3+e^(-n)).
Replace n with n+1 to get:
e^(-n-1) (2n+2+2/3-e^(-n-1)) < G(n+1) < e^(-n-1) (2n+2+2/3+e^(-n-1)).
(Sorry this isn't in tex!)
Now I will build up to F(n+1) from this, knowing that
F(n+1)=2n +2+ 2/3 - G(n+1) e^(n+1).
Ok, first multiply all sides of the inequality by -e^(n+1) to get this:
- (2n+2+2/3-e^(-n-1)) > -e^(n+1) G(n+1) > - (2n+2+2/3+e^(-n-1)); i.e.,
- (2n+2+2/3+e^(-n-1)) < -e^(n+1) G(n+1) < - (2n+2+2/3-e^(-n-1)).
Now, adding 2n+2+2/3 to this, I get
-e^(-n-1)) < 2n+2+2/3-e^(n+1) G(n+1) < e^(-n-1); i.e.,
-e^(-n-1)) < F(n+1) < e^(-n-1) ==> |F(n+1)|< (1/e)^(n+1).
phoenixthoth said:
Thank you for posting this wonderfully frustrating problem!
Here's what I've tried:
Trying to convert the limit of the sum part into an integral. If I could make the summands a function of k/n, then I'd be ok but I can't see how to do that.
Pretending that F(n) is the nth coefficient of a series. Proving the series converges would show that F(n) goes to 0. I haven't beaten this avenue to death yet; I haven't tried all viable convergence tests.
Seeing if Mathematica can do it. It can't. Haven't tried other CAS's. One thing about Mathematica is that apparently the numbers are very sensitive and its approximations can be waaaay off; at first I thought the limit diverged. But when I was a little more careful, it does seem to converge. I start having accuracy issues around n=150 to 200.
Trying to determine what that summand is the kth derivative of evaluated at 0. I think there's a post above about this.
And I forget what else. Oh yeah, I also briefly searched for a similar problem.
The current goose I'm following is this: proving, by induction, that for all n,
|F(n)| <= (1/e)^n .
This is true by mathematica for n up to 100. I started out with 1/2 instead of 1/e but then I tried 1/e and it still came out true for n up to 100. 1/e definitely seems like it will be better for this problem.