Finding order of an infinite series

[Moschops]

Homework Statement

I have been working on a truncated Fourier series. I have come up with a truncated series for cos (αx) and it matches my book, where in this case I'm letting x=π, and then I have shown as the book asks,

Truncated series = $F_{N}(π)$ = cos (απ) + $\frac{2α}{π} \sin{(απ)}$$\sum\limits_{k=N+1}^\infty$$\frac{1}{k^2 - α^2}$

I am then asked to show that this approximates to almost the same thing:

$F_{N}(π)$ = cos (απ) + $\frac{2α}{Nπ} \sin{(απ)}$

So I thought no worries, looks like I need to take that infinite sum there and simply show that it is of order 1/N.

Homework Equations

The Attempt at a Solution

The only way I know how to try that is to pretend the sum is actually an integral,

$\int \limits_{N+1}^\infty$$dk \frac{1}{k^2 - α^2}$, and I wound up with something like:

$\frac{\frac{αk}{α^2 - k^2} + arctan \frac{k}{α} }{2α^3}$

as the solution to the integral, with limits of k=infinity and k=N+1 in there for the definite integral (I can't figure out how to show that - I was lucky to manage what I have.

Anyway, having done that, it really doesn't look like it's of order 1/N to me. Am I heading the right way with this, or is there some other way to establish that it's of order 1/N?

This is the very last step in the question and I get the idea it's meant to be, if not trivial, at least quite an easy part.

 

[Mark44]

Homework Statement

I have been working on a truncated Fourier series. I have come up with a truncated series for cos (αx) and it matches my book, where in this case I'm letting x=π, and then I have shown as the book asks,

Truncated series = $F_{N}(π)$ = cos (απ) + $\frac{2α}{π} \sin{(απ)}$$\sum\limits_{k=N+1}^\infty$$\frac{1}{k^2 - α^2}$

I am then asked to show that this approximates to almost the same thing:

$F_{N}(π)$ = cos (απ) + $\frac{2α}{Nπ} \sin{(απ)}$

So I thought no worries, looks like I need to take that infinite sum there and simply show that it is of order 1/N.

Homework Equations

The Attempt at a Solution

The only way I know how to try that is to pretend the sum is actually an integral,

$\int \limits_{N+1}^\infty$$dk \frac{1}{k^2 - α^2}$, and I wound up with something like:

$\frac{\frac{αk}{α^2 - k^2} + arctan \frac{k}{α} }{2α^3}$

I don't know if your result is correct. An easier way to do the integration is to decompose the integrand using partial fractions.

$\frac{1}{k^2 - α^2} = \frac{A}{k - α} + \frac{B}{k + α}$
Solve the equation above for A and B and then integrate.

as the solution to the integral, with limits of k=infinity and k=N+1 in there for the definite integral (I can't figure out how to show that - I was lucky to manage what I have.

Anyway, having done that, it really doesn't look like it's of order 1/N to me. Am I heading the right way with this, or is there some other way to establish that it's of order 1/N?

This is the very last step in the question and I get the idea it's meant to be, if not trivial, at least quite an easy part.