# How do I determine if the series converges?

• Mar 12th 2008, 03:10 AM
KumaKuma
How do I determine if the series converges?
Its given that the series are as follows:

∑ (arctan k)/(1 + kČ)
k=1
• Mar 12th 2008, 03:36 AM
mr fantastic
Quote:

Originally Posted by KumaKuma
Its given that the series are as follows:

∑ (arctan k)/(1 + kČ)
k=1

$\displaystyle 0 < \frac{\pi}{4} < \frac{\arctan k}{1 + k^2} < \left( \frac{\pi}{2} \right) \, \frac{1}{k^2} < \frac{2}{k^2}$ for k > 0 ......
• Mar 12th 2008, 04:04 AM
KumaKuma
Quote:

Originally Posted by mr fantastic
$\displaystyle 0 < \frac{\pi}{4} < \frac{\arctan k}{1 + k^2} < \left( \frac{\pi}{2} \right) \, \frac{1}{k^2} < \frac{2}{k^2}$ for k > 0 ......

Sorry...im poor in calculus..not very clear with your provided answer, can you explain further??
• Mar 12th 2008, 05:17 AM
mr fantastic
Quote:

Originally Posted by KumaKuma
Sorry...im poor in calculus..not very clear with your provided answer, can you explain further??

There's no calculus involved here.

Do you understand how to use the basic comparison test?
• Mar 12th 2008, 05:41 AM
colby2152
Quote:

Originally Posted by mr fantastic
There's no calculus involved here.

Do you understand how to use the basic comparison test?

It is your best friend when proving the convergence of a series.
• Mar 12th 2008, 06:16 AM
KumaKuma
I did some searchings and readings about basic comparison test but i dont fully understand them.
Is the comparison between arctan k/1+kČ with Pi/4 and with (Pi/2)/(1/kČ) necessary?
Can i just compare in this manner?
0 < arctan k/ (1+kČ) < 2/kČ
I could not understand why that the numerator in the right equation has to be 2? can we use arctan k to replace the 2?
• Mar 12th 2008, 08:47 AM
KumaKuma
That was slightly confusing. I'll try to put them into proper notation.
What i meant was can i do the comparison without comparing $\displaystyle \frac{\pi}{4}$ and $\displaystyle \left( \frac{\pi}{2} \right) \, \frac{1}{k^2}$ ?
And can i compare it like this:
$\displaystyle 0 < \frac{\arctan k}{1 + k^2} < \frac{\arctan k}{k^2}$ instead of using $\displaystyle <\frac{2}{k^2}$ ? Why the numerator is chosen to be 2?
• Mar 12th 2008, 10:22 AM
Moo
Hello,

The factor is 2 because pi/2 is slightly inferior to 2. Actually, it's not a problem, we don't care about the constant factor. All we want to know is if we can compare the series with a Riemann series, which is the case, with a constant factor (not influencing the convergence) =)

arctan is always < pi/2 ;)
• Mar 13th 2008, 03:35 AM
KumaKuma
Ohh that cleared my doubts (Giggle)
So it will be like:
Since $\displaystyle \arctan k \leq \frac{\pi}{2}$

$\displaystyle 0 < \frac{\arctan k}{1 + k^2} < \frac{2}{k^2}$

$\displaystyle \sum_{k=1}^{\infty} \frac{2}{k^2} = 2\sum_{k=1}^{\infty} \frac{1}{k^2}$

Since $\displaystyle \frac{2}{k^2}$ converges, and thus $\displaystyle \frac{\arctan k}{1 + k^2}$ converges according to the basic comparison theorem.

Am I getting it right?
• Mar 13th 2008, 03:36 AM
mr fantastic
Quote:

Originally Posted by KumaKuma
Ohh that cleared my doubts (Giggle)
So it will be like:
Since $\displaystyle \arctan k \leq \frac{\pi}{2}$

$\displaystyle 0 < \frac{\arctan k}{1 + k^2} < \frac{2}{k^2}$

$\displaystyle \sum_{k=1}^{\infty} \frac{2}{k^2} = 2\sum_{k=1}^{\infty} \frac{1}{k^2}$

Since $\displaystyle \frac{2}{k^2}$ converges, and thus $\displaystyle \frac{\arctan k}{1 + k^2}$ converges according to the basic comparison theorem.

Am I getting it right?

(Clapping)
• Mar 13th 2008, 04:36 AM
KumaKuma
Ah a handclap, finally (Smile)
Thanks to all of you who have helped me! (Bigsmile)