show that the above series is converges
The series can be written in compact form as...
It is well known the limit...
... where is the 'Euler's constant'. Consequence of (3) is...
... 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...
Aw, man I was trying to figure out the function for it, glad I started 5 minutes ago
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
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 .