The exponential function #e^x# can be defined as a power series as: #e^x=sum_(n=0)^oo x^n/(n!)=1+x+x^2/(2!)+x^3/(3!)+...# Can you use this definition to evaluate #sum_(n=0)^(oo)((0.2)^n e^-0.2)/(n!)#?

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Answer 1 · 2017-04-25T01:19:37

#sum_(n=0)^(oo)((0.2)^n e^-0.2)/(n!) = 1#

We can write the sum as:

#sum_(n=0)^(oo)((0.2)^n e^-0.2)/(n!) = e^-0.2 \ sum_(n=0)^(oo)((0.2)^n )/(n!) \ \ .....(star)#

And using the above sum definition of #e^x# we have:

# sum_(n=0)^(oo)((0.2)^n )/(n!) = e^(0.2)#

Substituting this result into #(star)# we get:

# sum_(n=0)^(oo)((0.2)^n e^-0.2)/(n!) = e^-0.2 \ e^(0.2) #
# " " = e^0 #
# " " = 1 #