Evaluate the series or state that it diverges. \[ \sum_{k=1}^{\infty}\left[\frac{1}{7}\left(\frac{1}{6}\right)^{k}+\frac{2}{3}\left(\frac{1}{4}\right)^{k}\right] \]
Solution
Step 1 ：The given series is a sum of two geometric series. The sum of a geometric series can be calculated using the formula \(S = \frac{a}{1 - r}\), where \(a\) is the first term of the series and \(r\) is the common ratio.
Step 2 ：For the first series, \(a = \frac{1}{7}\) and \(r = \frac{1}{6}\). Substituting these values into the formula, we get the sum of the first series as \(S1 = \frac{1/7}{1 - 1/6} = 0.1714285714285714\).
Step 3 ：For the second series, \(a = \frac{2}{3}\) and \(r = \frac{1}{4}\). Substituting these values into the formula, we get the sum of the second series as \(S2 = \frac{2/3}{1 - 1/4} = 0.8888888888888888\).
Step 4 ：The sum of the given series is the sum of the sums of the two geometric series, i.e., \(S = S1 + S2 = 0.1714285714285714 + 0.8888888888888888 = 1.0603174603174603\).
Step 5 ：Final Answer: The sum of the series is \(\boxed{1.0603174603174603}\).
