How to find the sum of the series sum_"n=0"^oo(z^(4n)/((4n)!))n=0(z4n(4n)!) when |z| < 1 ?

1 Answer

sum_(k=0)^oo z^(4k)/(4k!)= (cosh(z)+cosh(i z) )/2k=0z4k4k!=cosh(z)+cosh(iz)2

Explanation:

We have

e^z = sum_(k=0)^oo z^k/(k!)ez=k=0zkk! and

e^-z = sum_(k=0)^oo (-1)^kz^k/(k!)ez=k=0(1)kzkk!

so

e^z+e^-z=2sum_(k=0)^oo z^(2k)/(2k!)ez+ez=2k=0z2k2k!

or

cosh(z)=sum_(k=0)^oo z^(2k)/(2k!)cosh(z)=k=0z2k2k!

now doing

cosh(z)+cosh(i z) = sum_(k=0)^oo z^(2k)/(2k!)+sum_(k=0)^oo (-1)^kz^(2k)/(2k!) = 2sum_(k=0)^oo z^(4k)/(4k!)cosh(z)+cosh(iz)=k=0z2k2k!+k=0(1)kz2k2k!=2k=0z4k4k!

Finally

sum_(k=0)^oo z^(4k)/(4k!)= (cosh(z)+cosh(i z) )/2k=0z4k4k!=cosh(z)+cosh(iz)2

or also

sum_(k=0)^oo z^(4k)/(4k!)= (cosh(z)+cos(z) )/2k=0z4k4k!=cosh(z)+cos(z)2