# 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!)#?

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

We can write the sum as:

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To evaluate ( \sum_{n=0}^{\infty} \frac{(0.2)^n e^{-0.2}}{n!} ), we can utilize the given power series expansion of ( e^x ).

First, let's rewrite the sum using the power series expansion of ( e^x ):

[ \sum_{n=0}^{\infty} \frac{(0.2)^n e^{-0.2}}{n!} = e^{-0.2} \sum_{n=0}^{\infty} \frac{(0.2)^n}{n!} ]

Now, we recognize that the summation ( \sum_{n=0}^{\infty} \frac{(0.2)^n}{n!} ) is the power series expansion of ( e^{0.2} ). Therefore:

[ \sum_{n=0}^{\infty} \frac{(0.2)^n e^{-0.2}}{n!} = e^{-0.2} \cdot e^{0.2} = e^{0} = 1 ]

So, ( \sum_{n=0}^{\infty} \frac{(0.2)^n e^{-0.2}}{n!} = 1 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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