# How do you use a power series to find the exact value of the sum of the series #1+2+4/(2!) +8/(3!) +16/(4!) + …# ?

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To find the exact value of the sum of the series (1 + 2 + \frac{4}{2!} + \frac{8}{3!} + \frac{16}{4!} + \ldots), we recognize that it resembles a geometric series with each term being a multiple of the previous term.

Let's denote the (n)th term of the series as (a_n). We observe:

[a_n = 2^{n-1} \cdot \frac{1}{(n-1)!}]

Now, to express this series as a power series, we can start by considering the exponential function (e^x), which has the power series representation:

[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots]

Comparing this with the series we have, we can make a substitution: (x = 2). This gives us:

[e^2 = 1 + 2 + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \ldots]

Now, multiply both sides by (e^{-2}):

[e^{-2} \cdot e^2 = e^{-2} \cdot (1 + 2 + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \ldots)]

[e^{2} \cdot e^{-2} = e^0 = 1]

[1 = e^{-2} + 2 \cdot e^{-2} + \frac{2^2}{2!} \cdot e^{-2} + \frac{2^3}{3!} \cdot e^{-2} + \frac{2^4}{4!} \cdot e^{-2} + \ldots]

This gives us:

[1 = e^{-2}(1 + 2 + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \ldots)]

[1 = e^{-2} \cdot \sum_{n=0}^\infty \frac{2^n}{n!}]

Now, recall that (e^x) has a Maclaurin series representation:

[e^x = \sum_{n=0}^\infty \frac{x^n}{n!}]

Substitute (x = -2) into this series:

[e^{-2} = \sum_{n=0}^\infty \frac{(-2)^n}{n!}]

Now, multiply both sides by (e^{-2}):

[e^{-2} \cdot e^{-2} = e^{-4} = \sum_{n=0}^\infty \frac{(-2)^n}{n!} \cdot e^{-2}]

[e^{-4} = \sum_{n=0}^\infty \frac{(-2)^n}{n!} \cdot e^{-2}]

[e^{-4} = \sum_{n=0}^\infty \frac{(-2)^n}{n!} \cdot \sum_{n=0}^\infty \frac{2^n}{n!}]

Now, equating the coefficient of (x^n) on both sides:

[\frac{1}{n!} = \sum_{k=0}^n \frac{(-2)^k}{k!} \cdot \frac{2^{n-k}}{(n-k)!}]

Finally, we're interested in the term where (n = 0, 1, 2, \ldots), so we can find the coefficient of (x^n) in the series expansion of (e^{-4}). This coefficient will give us the sum of the series (1 + 2 + \frac{4}{2!} + \frac{8}{3!} + \frac{16}{4!} + \ldots).

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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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