To evaluate the infinite sum
\[ \sum_{i=1}^{\infty} \frac{4}{7} \left( \frac{7}{6} \right)^{i-1}, \]
we can recognize that this is a geometric series.
The general form for the sum of an infinite geometric series \(\sum_{n=0}^{\infty} ar^n\) is given by:
\[ S = \frac{a}{1 - r} \]
where \(a\) is the first term of the series, and \(r\) is the common ratio, provided that \(|r| < 1\).
In our case, rewriting the sum starting from \(i = 1\) to \(i = \infty\), we can note that:
- \(a = \frac{4}{7}\) (this is the factor outside the summation).
- The factor \(\left( \frac{7}{6} \right)^{i-1}\) indicates that \(r = \frac{7}{6}\).
However, since the series starts from \(i=1\), we can express it as:
\[ \sum_{i=1}^{\infty} ar^{i-1} = a \sum_{n=0}^{\infty} r^{n}. \]
Here, by recognizing \(a = \frac{4}{7}\) and \(r = \frac{7}{6}\), we can calculate the sum. Notice the issue with convergence:
We need to check the convergence criterion which states that the infinite geometric series converges only if \(|r| < 1\). In this case,
\[ | \frac{7}{6} | = \frac{7}{6} > 1. \]
Since \(|r| > 1\), the series does not converge.
Thus, the answer is:
It does not converge.
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