show that the above series is converges(Rofl)

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- November 26th 2009, 08:53 PMXingyuanSeries converge

show that the above series is converges(Rofl) - November 26th 2009, 09:03 PMRockHard
Looks like somewhat are alternating series to me, find a function to which to fit the pattern of the terms you were given and simply use the alternating series test if you can

- November 26th 2009, 09:35 PMXingyuan
You probably mean this :

(Giggle) - November 26th 2009, 09:40 PMchisigma
The series can be written in compact form as...

(1)

... where...

(2)

It is well known the limit...

(3)

... where is the 'Euler's constant'. Consequence of (3) is...

(4)

... so that the (1) is an alternating decreasing terms series with the general term that in absolute value tends to 0 if . Therefore (1) is convergent...

Kind regards

- November 26th 2009, 09:52 PMRockHard
Aw, man I was trying to figure out the function for it, glad I started 5 minutes ago (Wink)

That is a new version of the alternating series I have not worked on yet so thank you for the question and thank you to chisigma for the answer.

Holy, that is a very new series in this form for me I would have never got natural log, how do you know when to apply such things chisigma - November 26th 2009, 11:26 PMmathaddict
HI

You can break these to find the general term .

1/1-1/3+1/5-1/7+... notice that the denominators are odd and the signs alternate so it would be

The other part :

(1)+(1+1/2)+(1+1/2+1/3)+... The pattern should be obvious here . Every subsequent fraction added has its denominator equal to the no of terms .

Putting them together :

Notice that as n approaches infinity , would approach 0 so that this series would eventually converge to 0 . - November 26th 2009, 11:31 PMRockHard
- November 27th 2009, 09:21 AMmathaddict