Integral test by comparison(Please verify my proof)

[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 $$a_n \geq a_{n+1}$$

Homework Equations

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

 

[PingPong]

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

 

[Hummingbird25]

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.