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Jul10-11, 09:22 PM
P: 621

The above series is convergent for abs[x]<1. Integration is allowed for such cases.

When abs[x]>1 we may proceed as follows:
Let y=1/x
Now, abs value of y is less than 1


Since y<1 , we may proceed exactly in the same manner and get the same
result preceded by a negative sign as expected.


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