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:

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

The exponential function e^xe^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!)+...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!)sum_(n=0)^(oo)((0.2)^n e^-0.2)/(n!)?