# Homework Help: Riemann Sum Proof

1. Dec 13, 2011

### PiRGood

1. The problem statement, all variables and given/known data
Prove that:
lim n->inf1/n*Ʃn-1k=0ekx/n

=

(ex-1)/x

x>0

2. Relevant equations

That was all the information provided. Any progress i have made is below. I didn't want to add any of that to this section because this is all speculation on my part so far.

3. The attempt at a solution
I've been at this for awhile now, i feel as though i am getting close. I think i have all the "pieces" but i cant seem to put them together to prove the above statement.
I know that the integral
01etxdt is important because it integrates to
(ex-1)/x
but i'm not sure how to connect the summation to the integral to the answer
I also have a feeling that the Theorem
is relevant. But i'm not positive.

Any help would be extremely appreciated!

2. Dec 13, 2011

### Dick

The thing your really need to notice to evaluate that sum is that e^(kx/n) is a geometric series. It's r^k where r=e^(x/n). There is a formula to sum a geometric series. Then, sure, your theorem tells you the limit as n->infinity of the sum is the integral. So now after you've simplified the sum try to find the limit.

Last edited: Dec 13, 2011
3. Dec 13, 2011

### PiRGood

If i recall correctly that formula is
I tried that but i hit a roadblock. I came up with 1-exn/n/ 1-ex/n. You can cancel the n's in the numerator's exponent of course. But that is where i hit my dead end. I couldn't manipulate it from there in any meaningful way, even using the 1/n factor in front of the summation.
The reason i abandoned that course is because my professor hinted that i should look in the chapter of my book regarding Riemann Sums. But i've yet to dig up anything useful besides the Theorem i quoted in the OP

4. Dec 13, 2011

### Dick

That's a good start! So you've got a 1-e^x part. Now look at the denominator, lim n*(1-e^(x/n)) as n->infinity. That's an infinity*0 form, right? That suggests you might want to write it as (1-e^(x/n))/(1/n). That's a 0/0 form and you can apply l'Hopital's rule.

Last edited: Dec 13, 2011
5. Dec 13, 2011

### PiRGood

But wouldn't taking the derivative of 1/n give us zero, because n is some constant approaching inf? So the derivative of (1-e^(x/n)) would be (xe^(x/n))/n and the derivative of 1/n would be 0, giving me ((xe^(x/n))/n )/0?

6. Dec 13, 2011

### Dick

Well, no. n isn't constant. It's a variable approaching infinity. x is the constant while you are taking the limit. You want to take d/dn of the numerator and denominator.

7. Dec 13, 2011

### PiRGood

So i am are applying L'Hôpital's rule to the fraction in the denominator of our function i.e applying it to (1-e^(x/n))/(1/n)
So i get -xe^(x/n)/(n^2) on top and -1/n^2 on the bottom which simplifies to:
xe^(x/n)?

8. Dec 13, 2011

### Dick

You missed a minus sign. But otherwise ok! Now take limit n->infinity of x*e^(x/n).

9. Dec 13, 2011

### PiRGood

as n approaches infinity x/n approaches zero, which means e^(x/n) approaches one which leaves me with x on the bottom!
Incredible, i never would have seen that. Thanks so much Dick! You are a lifesaver!

10. Dec 13, 2011

### Dick

Yeah, it does work out nicely and you are welcome. But so far you've got (1-e^x)/x, and you want to get (e^x-1)/x. Better find that lost minus sign before you call it a done deal.