Integral test by comparison(Please verify my proof)

• Hummingbird25
In summary, the conversation discusses the use of the integral test by comparison to prove that a_n \geq a_{n+1}. The proof involves showing that the integrand of a_{n+1} is always smaller than that of a_n on the interval (0,\pi). This is done by simplifying the difference between the two integrals and using the fact that the denominator on the negative half of the interval is larger than that of any other point on the positive half.
Hummingbird25
Integral test by comparison(Please look at my work)

Homework Statement

Looking at the Integral

$$a_n = \int_{0}^{\pi} \frac{sin(x)}{x+n\pi}$$

prove that $$a_n \geq a_{n+1}$$

The Attempt at a Solution

Proof

given the integral test of comparison and since a_n is convergent, then a_n will always be larger than a_(n+1), by comparisons test.

q.e.d.

Is this surficient? Or do I need to add something that they converge to different limit point?

Sincerely Yours
Maria

Last edited:
What can you say about the integrand of $a_{n+1}$ as compared to that of $a_{n}$ on the interval $(0,\pi)$?

Here is my now proof:

the difference between the two integrals, we seek to show:

$$\forall n\in\mathbb{N}:\int_0^\pi\left({\sin t\over t+n\pi}-{\sin t\over t+(n+1)\pi}\right)\,dt\ge 0$$

Common denominator:

$$=\int_0^\pi\left({\sin t((t+(n+1)\pi)-(t+n\pi))\over (t+n\pi)(t+(n+1)\pi)}\right)\,dt$$

$$=\pi\int_0^\pi\left({\sin t\over (t^2+(2n+1)\pi t+(n^2+n)\pi)}\right)\,dt$$

From here we can use the fact that the denominator on the half of the interval where sinus is negative is larger than the denominator of each and every other corresponding point on the other half of the interval, so the whole integral must be positive.

q.e.d.

How does it look now?

Sincerely Maria.

Last edited:

1. What is the integral test by comparison?

The integral test by comparison is a method used to determine the convergence or divergence of an infinite series by comparing it to an integral function. It is based on the comparison test for series, which states that if the terms of a series are always less than the terms of another convergent series, then the original series must also converge.

2. How does the integral test by comparison work?

The integral test by comparison works by comparing the terms of an infinite series to the terms of an integral function. If the integral function is convergent, then the series must also converge. If the integral function is divergent, then the series must also diverge. This is because the integral function provides a bound for the series, which helps determine its convergence or divergence.

3. What are the conditions for using the integral test by comparison?

The conditions for using the integral test by comparison are that the function f(x) must be positive, continuous, and decreasing on the interval [1, ∞), and the series must have non-negative terms. Additionally, the series and the integral function must have the same behavior at infinity, meaning they both either converge or diverge.

4. Can the integral test by comparison be used to determine the sum of a series?

No, the integral test by comparison can only be used to determine the convergence or divergence of a series. It does not provide a method for finding the actual sum of a series. Other methods such as the geometric series or telescoping series must be used to find the sum of a series.

5. Are there any limitations to the integral test by comparison?

Yes, the integral test by comparison only works on series with non-negative terms. It also does not work on series with oscillating or alternating terms, such as the alternating harmonic series. In these cases, other methods such as the alternating series test must be used to determine the convergence or divergence of the series.

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