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→∞(1−z2n2)n=1, I use the binomial expansion of the series and write it down limn→∞∑nk=0z2kn2kn!n−k!k! and further expand the combinatorial term. How do I prove that the limit of this is 1 ?
Answer
Here is an approach.
enln(1−z2/n2)=en(−z2/n2+O(z4/n4))⟶n→∞1.
We used the Taylor series
ln(1−t)=−t+O(t2).
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