Find the sum of the following series
(i) $\sum_{n=1}^{\infty}\left[(-0.2)^{n}+(0.6)^{n-1}\right]$

Step 1 ：We can rewrite this as two separate series

Step 2 ：\[\sum_{n=1}^\infty \left[(-0.2)^n + (0.6)^{n-1}\right] = \sum_{n=1}^\infty (-0.2)^n + \sum_{n=1}^\infty (0.6)^{n-1}\]

Step 3 ：The first series is a geometric series with first term -0.2 and common ratio -0.2, so we have

Step 4 ：\[\sum_{n=1}^\infty (-0.2)^n = \frac{-0.2}{1-(-0.2)} = -0.2\]

Step 5 ：The second series is also a geometric series, but the first term is 0.6 and the common ratio is 0.6. However, the index starts from n-1 instead of n. To make it start from n, we can rewrite the series as

Step 6 ：\[\sum_{n=1}^\infty (0.6)^{n-1} = \sum_{n=0}^\infty (0.6)^n\]

Step 7 ：Now, we can calculate the sum of the second series

Step 8 ：\[\sum_{n=0}^\infty (0.6)^n = \frac{0.6}{1-0.6} = 1.5\]

Step 9 ：Hence, the sum of the original series is

Step 10 ：\[\sum_{n=1}^\infty \left[(-0.2)^n + (0.6)^{n-1}\right] = -0.2 + 1.5 = \boxed{1.3}\]