Its given that the series are as follows:

∞

∑ (arctan k)/(1 + kČ)

k=1

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- Mar 12th 2008, 04:10 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, 04:36 AMmr fantastic
- Mar 12th 2008, 05:04 AMKumaKuma
- Mar 12th 2008, 06:17 AMmr fantastic
- Mar 12th 2008, 06:41 AMcolby2152
- Mar 12th 2008, 07:16 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, 09:47 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 and ?

And can i compare it like this:

instead of using ? Why the numerator is chosen to be 2? - Mar 12th 2008, 11:22 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, 04:35 AMKumaKuma
Ohh that cleared my doubts (Giggle)

So it will be like:

Since

Since converges, and thus converges according to the basic comparison theorem.

Am I getting it right? - Mar 13th 2008, 04:36 AMmr fantastic
- Mar 13th 2008, 05:36 AMKumaKuma
Ah a handclap, finally (Smile)

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