How do you prove that the limit #n to oo sum e^(-n/10) cos(n/10)#, for n from 0 to n, is 10.5053, nearly.?

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Answer 1 · 2016-12-03T19:51:55

See below.

#e^(-k/10)(cos(-k/10)+i sin(-k/10))=e^(-k/10)e^(-ik/10) = e^(-((i+1)/10)k)#

so

#sum_(k=0)^ooe^(-k/10)cos(k/10) = R_e(sum_(k=0)^oo e^(-((i+1)/10)k))#

but #abs(e^(-((i+1)/10)k))=e^(-k/10) < 1# so

#sum_(k=0)^oo e^(-((i+1)/10)k) = (0-1)/(e^(-(i+1)/10)-1)# so

#1/(1-(e^(-(i+1)/10)))=1/(2 - (2 sinh[1/10])/(e^(1/10) - Cos(1/10)))-i((e^(1/10) Sin(1/10))/(1 + e^(1/5) - 2 e^(1/10) Cos(1/10)))#

and

#R_e(1/(1-(e^(-(i+1)/10))))=1/(2 - (2 sinh[1/10])/(e^(1/10) - Cos(1/10)))=5.508336109787682#