Sums of series for Riemann integrals

In summary, for the given sums related to Riemann integrals, you can use the formulas for the sum of arithmetic and geometric series to find the sums of the first n terms. For the third sum, using the fact that sin(k*pi/2n) = Im(e^i(k*pi/2n)), you can get the sum as 0.
  • #1
ThomasR
2
0
Hello

I'm having some difficulty in finding sums which relate to Riemann integrals.

The first one seems pretty simple.. a finite calculation of what would otherwise be the harmonic series i.e. 1/k from k=n to k=(2n-1). I can't see an easy way of finding a formula in terms of n, however.

The second one is similar.. sum of k/[n(k+n)] from k=1 to k=n. Again, completely stumped here as to how to get it into a nice format.

Finally, I need to sum from k=1 to k=(n-1) the values of sin(k*pi/2n). I have a similar result for cos(k*pi/2n) and it mentions using trigonometric/geometric series (which I presume means using the fact that sin(k*pi/2n) = Im(e^i(k*pi/2n)) but am unsure where to go from here.

Help on any of the above would be appreciated .. or just some general approach.

Cheers
Tom
 
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  • #2
For the first one, you can use the formula for the sum of the first n terms of an arithmetic series to solve for the sum of the harmonic series. The formula is S = (n/2)*[2*a+(n-1)*d], where a is the first term, d is the common difference, and n is the number of terms. In this case, a = 1/n and d = 1/(n-1), so the sum of the harmonic series would be: S = (n/2)*[2*(1/n) + (n-1)*(1/(n-1))], which simplifies to S = n/2 + (n-1)/2. For the second one, you can use the formula for the sum of the first n terms of a geometric series to solve for the sum of the series k/[n(k+n)]. The formula is S = a*(1-(r^n))/(1-r), where a is the first term, r is the common ratio, and n is the number of terms. In this case, a = 1/n and r = -1/n, so the sum of the series would be S = (1/n)*[1-(-(1/n)^n)]/(1-(-1/n)), which simplifies to S = n. For the third one, you can use the fact that sin(k*pi/2n) = Im(e^i(k*pi/2n)) to solve for the sum. Using the formula for the sum of the first n terms of a geometric series, you can get the sum of the series as S = Im(e^i(pi/2n))*(1-(e^i(pi/n))^n)/(1-e^i(pi/n)), which simplifies to S = 0.
 

1. What is a Riemann integral?

A Riemann integral is a mathematical concept that represents the area under a curve on a graph. It is used to calculate the total value of a function over a specific interval.

2. How is a Riemann integral calculated?

A Riemann integral is calculated by dividing the area under the curve into smaller rectangles and finding the sum of their areas. This is done by taking the limit as the width of the rectangles approaches zero and the number of rectangles approaches infinity.

3. What is the significance of Riemann integrals?

Riemann integrals have significant applications in mathematics, physics, and engineering. They are used to solve problems related to calculating the total value of a function, finding the area under a curve, and evaluating complex equations.

4. Can Riemann integrals be used to integrate any function?

No, Riemann integrals can only be used to integrate functions that are continuous and defined over a finite interval. They cannot be used for functions with discontinuities or infinite intervals.

5. What is the relationship between Riemann integrals and series?

Riemann integrals and series are closely related as the value of a Riemann integral can be expressed as the limit of a series. This is known as the Riemann series theorem and it allows for the evaluation of difficult integrals using series methods.

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