# Finding order of an infinite series

• Moschops
In summary, the conversation discusses a truncated Fourier series and how it approximates to a similar series. The speaker is trying to show that the infinite sum in the series is of order 1/N, and they attempt to do so by integrating the sum. However, it is suggested that an easier method would be to decompose the integrand using partial fractions. The speaker is unsure if their result is correct and is looking for confirmation.
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.

## 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.

Last edited:
Moschops said:

## 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.

## 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.
Moschops said:
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.

## 1. What is an infinite series?

An infinite series is a sum of an infinite number of terms. It is written in the form of a1 + a2 + a3 + ... + an, where a1, a2, a3, ..., an are the terms of the series.

## 2. How do you find the order of an infinite series?

The order of an infinite series is determined by the highest power of the variable in the series. For example, if the series is written as a1x + a2x^2 + a3x^3 + ..., the order would be 3.

## 3. What is the purpose of finding the order of an infinite series?

Finding the order of an infinite series helps determine the convergence or divergence of the series. It also helps in evaluating the series using various mathematical techniques.

## 4. What is the difference between a convergent and divergent series?

A convergent series is a series whose terms approach a finite value as you add more terms. In other words, the sum of the series has a finite value. A divergent series is a series whose terms do not approach a finite value and the sum of the series is infinite.

## 5. How do you determine the convergence or divergence of an infinite series?

There are various tests that can be used to determine the convergence or divergence of an infinite series, such as the ratio test, the root test, and the integral test. These tests compare the series to known convergent or divergent series and can help determine the behavior of the infinite series.

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