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Unformatted text preview: radius of convergence? Z ln(1-t ) t dt Solution : Note that d dt ln(1-t ) =-1 1-t , which is the sum of the geometric series ∑ ∞ n =0-t n . Thus, we have that ln(1-t ) = Z ∞ X n =0-t n dt = ∞ X n =0 Z-t n dt = ∞ X n =0-1 n + 1 t n +1 + C. Plugging in t = 0 yields ln(1) = 0 = ∑ 0 + C , so that C = 0 and ln(1-t ) = ∑ ∞ n =0-1 n +1 t n +1 . Finally, dividing by t yields ln(1-t ) t = ∞ X n =0-1 n + 1 t n . We can integrate this to get Z ln(1-t ) t dt = Z ∞ X n =0-1 n + 1 t n dt = ∞ X n =0 Z-1 n + 1 t n dt = ∞ X n =0-1 ( n + 1) 2 t n +1 + C. The radius of convergence of the very ﬁrst series we created is 1 (a geometric series with ratio t ), and none of our steps do anything to change that. Thus, our ﬁnal radius of convergence is also 1....
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- Spring '08
- Power Series, power series representation, n=0