sum_(n=-1)^infinity(-1)^(n+1) (1/3)^n = 9/4~~2.25...
How do you find that out? Is there a formula?
To find the sum of the series \(\sum_{n=-1}^{\infty} (-1)^{n+1} \left(\frac{1}{3}\right)^n\), we can use the formula for the sum of an infinite geometric series.
The formula for the sum of an infinite geometric series is given by:
\[S = \frac{a}{1 - r}\]
Where:
- \(S\) is the sum of the series,
- \(a\) is the first term of the series, and
- \(r\) is the common ratio between consecutive terms.
Let's apply this formula to the given series:
First, we need to identify \(a\) and \(r\).
The first term, \(a\), can be calculated by substituting \(n = -1\) into the series expression:
\(\left(\frac{1}{3}\right)^{-1} = 3\)
The common ratio, \(r\), can be obtained by taking the ratio of any term to its previous term. So, let's consider \(n = 0\) and \(n = -1\) terms:
\(\frac{\left(\frac{1}{3}\right)^0}{\left(\frac{1}{3}\right)^{-1}} = \frac{1}{3}\times3 = 1\)
Now, we have \(a = 3\) and \(r = 1\). Plugging these values into the formula, we get:
\[S = \frac{3}{1 - 1} = \frac{3}{0}\]
However, we can see that the denominator is zero, which means the formula doesn't work in this case. This occurs when the common ratio \(r\) is equal to 1. In such cases, the formula for a sum of an infinite geometric series is not applicable.
To determine the sum in this case, we can rewrite the series as a finite sum and then take the limit as the number of terms approaches infinity.
Let's rewrite the series by splitting it into two parts:
\[
\sum_{n=-1}^{-1} (-1)^{n+1} \left(\frac{1}{3}\right)^n + \sum_{n=0}^{\infty} (-1)^{n+1} \left(\frac{1}{3}\right)^n
\]
The first part, \(\sum_{n=-1}^{-1} (-1)^{n+1} \left(\frac{1}{3}\right)^n\), has only one term:
\((-1)^{-1+1} \left(\frac{1}{3}\right)^{-1} = -3\)
The second part, \(\sum_{n=0}^{\infty} (-1)^{n+1} \left(\frac{1}{3}\right)^n\), can be computed using the formula for the sum of an infinite geometric series since it now starts from \(n = 0\):
\[
\sum_{n=0}^{\infty} (-1)^{n+1} \left(\frac{1}{3}\right)^n = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}
\]
Now, we sum up the two parts:
\[-3 + \frac{3}{2} = \frac{-6 + 3}{2} = -\frac{3}{2} = \frac{9}{4}\]
Therefore, the sum of the series is \(\frac{9}{4}\), which is approximately 2.25.