What is the radius of convergence of sum_1^oo e^(nx) / 2^(2n)?∞∑1enx22n??

1 Answer
Cesareo R.
Jul 18, 2016

x < log_e 4x<loge4

Explanation:

S(x)=sum_1^oo e^(nx) / 2^(2n)=sum_1^oo(e^x/4)^nS(x)=∞∑1enx22n=∞∑1(ex4)n

this series converges for e^x/4 <1ex4<1 giving as result

S(x) = 1/(1-e^x/4) = 4/(4-e^x), forall x | e^x<4->x < log_e 4S(x)=11−ex4=44−ex,∀x∣ex<4→x<loge4

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