Its given that the series are as follows:

∞

∑ (arctan k)/(1 + kČ)

k=1

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- Mar 12th 2008, 03:10 AMKumaKumaHow 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 AMmr fantastic
- Mar 12th 2008, 04:04 AMKumaKuma
- Mar 12th 2008, 05:17 AMmr fantastic
- Mar 12th 2008, 05:41 AMcolby2152
- Mar 12th 2008, 06:16 AMKumaKuma
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 AMKumaKuma
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 AMMoo
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 AMKumaKuma
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 AMmr fantastic
- Mar 13th 2008, 04:36 AMKumaKuma
Ah a handclap, finally (Smile)

Thanks to all of you who have helped me! (Bigsmile)