# Math Help - How do I determine if the series converges?

1. ## How do I determine if the series converges?

Its given that the series are as follows:

∑ (arctan k)/(1 + k²)
k=1

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

∑ (arctan k)/(1 + k²)
k=1
$0 < \frac{\pi}{4} < \frac{\arctan k}{1 + k^2} < \left( \frac{\pi}{2} \right) \, \frac{1}{k^2} < \frac{2}{k^2}$ for k > 0 ......

3. Originally Posted by mr fantastic
$0 < \frac{\pi}{4} < \frac{\arctan k}{1 + k^2} < \left( \frac{\pi}{2} \right) \, \frac{1}{k^2} < \frac{2}{k^2}$ for k > 0 ......
Sorry...im poor in calculus..not very clear with your provided answer, can you explain further??

4. 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?

5. 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.

6. I did some searchings and readings about basic comparison test but i dont fully understand them.
Is the comparison between arctan k/1+k² with Pi/4 and with (Pi/2)/(1/k²) necessary?
Can i just compare in this manner?
0 < arctan k/ (1+k²) < 2/k²
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?

7. That was slightly confusing. I'll try to put them into proper notation.
What i meant was can i do the comparison without comparing $\frac{\pi}{4}$ and $\left( \frac{\pi}{2} \right) \, \frac{1}{k^2}$ ?
And can i compare it like this:
$0 < \frac{\arctan k}{1 + k^2} < \frac{\arctan k}{k^2}$ instead of using $<\frac{2}{k^2}$ ? Why the numerator is chosen to be 2?

8. 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

9. Ohh that cleared my doubts
So it will be like:
Since $\arctan k \leq \frac{\pi}{2}$

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

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

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

Am I getting it right?

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

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

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

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

Am I getting it right?

11. Ah a handclap, finally
Thanks to all of you who have helped me!