a. For the Ratio Test, we need to examine the ratio of (k+1)-th coefficient to k-th coefficient, here it is
(1/((k+1)!)) /(1/(k!)) = (k!)/((k+1)!) = 1/(k+1).
The limit of this ratio is 0, therefore the power series converges everywhere (and there are no endpoints to check).
b. To determine the function to which the series converges, recall the definition of the Taylor series (with the center at x=0 ). For a function f(x) its Taylor series is sum_(k=0)^oo f^(k)(0) x^k/k!
Our series is sum_(k=0)^oo x^(k+1)/(k!) = x sum_(k=0)^oo x^k/(k!) = x e^x,
because (e^x)^((k)) = e^x and (e^x)^((k))(0) = 1.
http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/PowerSeries/
http://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries.aspx
Thursday, February 23, 2012
Part a Using Maple find and show the interval and radius of convergence of this series sum_(k=0)^oox^(k+1)/(k!) use ratio test and test the endpoints. Part b to what function does this series converge. xe^x is apparently correct but show why.
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