Its given that the series are as follows:
∞
∑ (arctan k)/(1 + kČ)
k=1
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?
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?
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
Ohh that cleared my doubts
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?