let x=3 therefore 1+3+3^2/2!+3^3/3!+3^4/4!+⋯+3^n/n!+ ... becomes: 1 + x + x^2/2! + x^3/3! + ... + x^n/n! We know that the inf SUM (n=0) x^n/n! = e^x
then you need to replace x so the answer should be e^3
Same Idea for this one let x=2 1 - x^2/2! + x^4/4! + x^6/6! +....+ [(-1)^n * x^(2n)/(2k!)]
We all know that the inf SUM (n=0) [(-1)^n * x^(2n)/(2k!)] = cos x
