Solve the equation of ? {n,0,infinity} (-4)^-n / ((2n)!)

Question

∑ {n,0,infinity} (-4)^-n / ((2n)!)

 

Can somebody give a step by step solution. 

The answer should be cos(1/2)

 

Thanks in advance!

Answer 1

 

Using the series expansion, e^(i/2)  =∑( k,0,infinity}(i/2)^k /k!  

 

Note that the real part consists of only the even  k

 

Re(e^(i/2)) = ∑(n,0,infinity}(i/2)^2n /(2n! ) =  ∑(n,0,infinity} (i^2/ 2^2) ^n / (2n! )

                                     = ∑(n,0,infinity} (-1/4)^n / (2n! )

                                     = ∑(n,0,infinity} (-4^-1)^n / (2n! ) .

                                     = ∑(n,0,infinity} (-4)^-n / (2n! )  .. using (x^a)^b = x ^(ab)

But

Re[e^(i/2) ] =  Re[cos(1/2) + i sin(1/2)] --- using euler's theorem

                   = cos(1/2) 

 

Therefore,  ∑(n,0,infinity} (-4)^-n / (2n! ) = cos(1/2)