How do you use a power series to find the exact value of the sum of the series #1+e+e^2 +e^3 +e^4 + …# ?
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To find the sum of the series (1 + e + e^2 + e^3 + e^4 + \ldots), we recognize that this is a geometric series with first term (a = 1) and common ratio (r = e).
The sum of an infinite geometric series is given by the formula:
[S = \frac{a}{1 - r}]
Substitute (a = 1) and (r = e) into the formula:
[S = \frac{1}{1 - e}]
However, this is not a power series.
To express this sum as a power series, we can rewrite it using a power series expansion. One way to do this is to use the Maclaurin series expansion for (e^x), which is:
[e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}]
Substitute (x = 1) into the series:
[e = \sum_{n=0}^{\infty} \frac{1^n}{n!} = \sum_{n=0}^{\infty} \frac{1}{n!}]
So, the sum of the series (1 + e + e^2 + e^3 + e^4 + \ldots) is:
[S = 1 + e + e^2 + e^3 + e^4 + \ldots = 1 + \left(\sum_{n=0}^{\infty} \frac{1}{n!}\right) + \left(\sum_{n=0}^{\infty} \frac{1}{n!}\right)^2 + \left(\sum_{n=0}^{\infty} \frac{1}{n!}\right)^3 + \ldots]
This is a series of series, but it can be simplified using the properties of power series. So the sum (S) is a complicated expression involving infinite sums of factorials.
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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.
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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