Tuesday, August 9, 2016

calculus - limit to prove exponential definition for all values of z



In order to prove that the relation limn(1+zn)n=exp(z) holds for all values of z, I first prove it for positive z by showing the lower and upper bound as taylor series for exp(z). To extend the proof of this relation to hold for negative z, I need to prove limn(1z2n2)n=1, I use the binomial expansion of the series and write it down limnnk=0z2kn2kn!nk!k! and further expand the combinatorial term. How do I prove that the limit of this is 1 ?


Answer



Here is an approach.




enln(1z2/n2)=en(z2/n2+O(z4/n4))n1.





We used the Taylor series




ln(1t)=t+O(t2).



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