evaluation of limit as a sum (1 Viewer)

[juantheron]

 

[Sy123]

 

[RealiseNothing]

It's in the form

Just integrate it now as this is a Riemann sum isn't it?

 

[RealiseNothing]

So

If you integrate this you get which agrees with Sy's answer.

 

[Sy123]

It's in the form

Just integrate it now as this is a Riemann sum isn't it?

So

If you integrate this you get which agrees with Sy's answer.

Yep its just a Riemann sum, I was just disguising my solution to make it HSC worthy

 

[mathing]

This is an interesting problem, however I am not sure the answers above rigorously prove it, as an integral is defined to be the limit between the upper and lower Riemann sums, and here we are technically asked to evaluate an infinite sum not an integral. Maybe the best we can do without evaluating the sum directly is just recognize that it is approximating the area of the quarter circle with higher precision as n increases. This should be all that would be required in any 4U exam.

 

[Carrotsticks]

We are asked to evaluate an infinite sum, but this infinite sum is equivalent to the integral above, as per the definition of the Riemann integral being the limiting case of the Riemann Sum. Hence by finding the integral, we find the infinite sum.

Sent from my SM-N9005 using Tapatalk

 

[mathing]

Good point, now that you mention it I do remember that notation from high school books. However I think that is still the non-rigorous way of defining the integral. It may be standard practice for HSC problems, I am not sure.

To see more about what I was talking about look at page 3 of https://www.math.ucdavis.edu/~hunter/m125b/ch1.pdf you can see the upper and lower Riemann sums used. If we just say it's the sum is the integral we are missing the step of proving that. What usually happens in the proofs is to be able to evaluate the sums and show they are the same, but what happened here was to say the sum is the integral to evaluate the sum.

 

[Carrotsticks]

We used the Riemann integral, which has 'inbuilt' (for this function) the upper and lower limits being equal to each other.

We could take the upper/lower limit blah blah but we will just end up with the Riemann integral.

Also note that the question is asking for the limit to be computed, not to be proven. If the question was asking for the limit to be proven carefully, then your idea of the approach may be more appropriate.

 

[seanieg89]

Good point, now that you mention it I do remember that notation from high school books. However I think that is still the non-rigorous way of defining the integral. It may be standard practice for HSC problems, I am not sure.

To see more about what I was talking about look at page 3 of https://www.math.ucdavis.edu/~hunter/m125b/ch1.pdf you can see the upper and lower Riemann sums used. If we just say it's the sum is the integral we are missing the step of proving that. What usually happens in the proofs is to be able to evaluate the sums and show they are the same, but what happened here was to say the sum is the integral to evaluate the sum.

The function is continuous, continuous functions are Riemann integrable over compact intervals. So the integral exists, and any Riemann sum will converge to it.

One might claim that we do not prove that continuous functions are integrable in high school, but by the same token we don't properly define integrals, limits, or even real numbers in high school. The HSC maths courses definitely aren't meant to be rigorous courses in analysis, so it is silly to apply quite the same standards of rigour.

The argument is perfectly valid (even if the vast majority of high school students won't understand why) and convincing to me.