How should I solve this problem?

[nombreuso]

Calculate integral from 1 to 0 of e^x dx by using upper Riemann sums. Hints: The sum is a geometric progression. You will need the limit limn→∞ n(e 1/n−1). This can be evaluated putting h = 1/n and relating the limit to the derivative of e x at x = 0. From the the limit part on I don't think I will have any problem.

 

[Deleted member 4993]

Calculate integral from 1 to 0 of e^x dx by using upper Riemann sums. Hints: The sum is a geometric progression. You will need the limit limn→∞ n(e 1/n−1). This can be evaluated putting h = 1/n and relating the limit to the derivative of e x at x = 0. From the the limit part on I don't think I will have any problem.

Please check your post for correctness.

the limit limn→∞ n(e 1/n−1)

That does not quite make sense.

Is it:

\(\displaystyle \lim_{n \to \infty}\left(n * e^{\left[\frac{1}{n}\right]}- 1\right) \) .............................. or something else

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[nombreuso]

Please check your post for correctness.


That does not quite make sense.

Is it:

\(\displaystyle \lim_{n \to \infty}\left(n * e^{\left[\frac{1}{n}\right]}- 1\right) \) .............................. or something else

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

Sorry, it didn't paste correctly. I meant lim as n approaches ∞ of n(e^(1/n) - 1)
I rewrote the integral from 0 to 1 of e^x dx as lim as n approaches 0 of 1/n(e^(0/n)+e^(1/n)+e^(2/n)+e^(3/n)+...+e^(n-1/n)+e^(n/n)

 

[Steven G]

As n->oo, n->oo and e^1/n->1. So as n->oo, n*e^1/n->oo. Subtracting 1 does not change that the limit is oo

 

[Deleted member 4993]

@Jomo

Herr Gauss will be extremely disappointed at your sudden transformation!!!!

 

[Cubist]

@Jomo

Herr Gauss will be extremely disappointed at your sudden transformation!!!!

One could say that @Jomo 's profile has evolved!
Did you need to click the deGauss button to remove the last pic? (Terrible jokes, sorry!) ?

It's a great choice, and it certainly continues to give an air of authority!

 

[Steven G]

I figured that my picture should reflect my signature.

 

[topsquark]

I figured that my picture should reflect my signature.

I haven't seen your signature. Do you write like you are at sea on the Beagle?

-Dan

 

[Deleted member 4993]

I figured that my picture should reflect my signature.

Is that a portrait of a blind man or a black cat?

 

[skeeter]

I meant lim as n approaches ∞ of n(e^(1/n) - 1)

[MATH]\lim_{n \to \infty} \dfrac{e^{\frac{1}{n}} - 1}{\frac{1}{n}}[/MATH]
L'Hopital ...

[MATH]\lim_{n \to \infty} \dfrac{e^{\frac{1}{n}} \cdot (\cancel{-\frac{1}{n^2}})}{\cancel{-\frac{1}{n^2}}} = 1[/MATH]

 

[lookagain]

As n->oo, n->oo and ...

Did you intend something else here instead where I put it in the bold text?
This looks to be a redundancy.

Also, regarding post #3, it is an indeterminate form of oo*0.
It is not equivalent to the lim as n --> oo of n*e^(1/n) - 1. It's the
lim as n --> oo of n[e^(1/n) - 1], which is also the lim as n --> oo of [n*e^(1/n) - n].

 

[Cubist]

L'Hopital ...

[MATH]\lim_{n \to \infty} \dfrac{e^{\frac{1}{n}} \cdot (\cancel{-\frac{1}{n^2}})}{\cancel{-\frac{1}{n^2}}} = 1[/MATH]

OP, alternatively you can use Skeeter's suggested method of L'Hopital after performing the suggested "h = 1/n" substitution. Obviously n=1/h, and you calculate the limit as h approaches 0. You obtain the same result but the derivatives are slightly easier.

 

[Cubist]

I rewrote the integral from 0 to 1 of e^x dx as lim as n approaches 0 of 1/n(e^(0/n)+e^(1/n)+e^(2/n)+e^(3/n)+...+e^(n-1/n)+e^(n/n)

You're close. However, since this is an upper Riemann sum, S, then e^(0/n) isn't the first term since it's the lowest value of the first subdivision and you're after the highest value. As a sum, I think you're after...

[math] S=\frac{1}{n} \sum_{p=1}^{n}{\mathrm{e}^{\frac{p}{n}}} [/math]
Can you simplify this/ remove the summation?

 

[Steven G]

Is that a portrait of a blind man or a black cat?

No, it is a picture of Darwin.

 

[Steven G]

Did you intend something else here instead where I put it in the bold text?
This looks to be a redundancy.

Also, regarding post #3, it is an indeterminate form of oo*0.
It is not equivalent to the lim as n --> oo of n*e^(1/n) - 1. It's the
lim as n --> oo of n[e^(1/n) - 1], which is also the lim as n --> oo of [n*e^(1/n) - n].

The original post did not have the n factored out. That however was changed at some point. I failed to read the memo.