Prove Riemann Sum: (ex-1)/x for x > 0

[PiRGood]

Homework Statement

Prove that:
lim _n->inf1/n*Ʃ^n-1_k=0e^kx/n

=

(e^x-1)/x

x>0

Homework 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.

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 can't seem to put them together to prove the above statement.
I know that the integral
₀∫¹e^txdt is important because it integrates to
(e^x-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

Suppose that f is defined on [a,b] and that lim_||P||->0Rp(f) exists. Then f is integrable on [a,b] and
_a∫^bf = lim_||P||->0Rp(f)

is relevant. But I'm not positive.

Any help would be extremely appreciated!

 

[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.

 

[PiRGood]

If i recall correctly that formula is

1-rⁿ/ 1-r

I tried that but i hit a roadblock. I came up with 1-e^xn/n/ 1-e^x/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

 

[Dick]

If i recall correctly that formula isI tried that but i hit a roadblock. I came up with 1-e^xn/n/ 1-e^x/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

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.

 

[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?

 

[Dick]

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?

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.

 

[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)?

 

[Dick]

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)?

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

 

[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!

 

[Dick]

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!

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.