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Unformatted text preview: ln(1t ) = Z1 1t dt = Z ∞ X n =0t n dt = ∞ X n =0 Zt n dt = ∞ X n =0t n +1 n + 1 + C. Plugging in t = 0 gives 0 = ln(10) = ∞ X n =0 0 + C, so that C = 0. Putting it all together, we’ve so far discovered that ln(1t ) = ∞ X n =0t n +1 n + 1 . Finally, multiplying by t and integrating yields Z t ln(1t ) dt = Z t ∞ X n =0t n +1 n + 1 dt = Z ∞ X n =0t n +2 n + 1 dt = ∞ X n =0 Zt n +2 n + 1 dt = ∞ X n =0t n +3 ( n + 1)( n + 3) + C. Finally, since the radius of convergence of our ﬁrst series ( ∑ ∞ n =0t n ) was 1, and we only changed it in ways (integration, multiplication by t ) that don’t aﬀect the radius of convergence, our ﬁnal radius is also R = 1. Alternatively, you can check this using Ratio Test if you are so inclined. I’m not....
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 Spring '08
 REMPE
 Maclaurin Series, Taylor Series, Mathematical Series, lim, Radius of convergence

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