The given sequence can be written as:

$S = (a + b) + (a^2 + ab + b^2) + (a^3 + a^2b + ab^2 + b^3) + \dots$

Step 1: Expressing the $n$-th term

For the $n$-th term of the sequence, the general term is the sum of all terms of the form $a^i b^j$ such that $i + j = n$. This can be written as: $T_n = \sum_{i=0}^n a^i b^{n-i}.$

Step 2: Summing the infinite series

The infinite sum can be written as: $S = \sum_{n=1}^\infty \sum_{i=0}^n a^i b^{n-i}.$

Now, let's rearrange the terms. Each pair $(i, j)$ appears exactly once, where $i + j \geq 1$. Hence, we can express the sequence as: $S = \sum_{i=0}^\infty \sum_{j=0}^\infty a^i b^j - 1,$ where we subtract 1 to account for the double-counting of the constant term $1$ (when $i = 0, j = 0$).

Step 3: Evaluate the geometric sums

The double summation can be evaluated by recognizing it as a product of two geometric series. Since $|a| < 1$ and $|b| < 1$, we have: $\sum_{i=0}^\infty a^i = \frac{1}{1-a} \quad \text{and} \quad \sum_{j=0}^\infty b^j = \frac{1}{1-b}.$

Thus: $\sum_{i=0}^\infty \sum_{j=0}^\infty a^i b^j = \frac{1}{(1-a)(1-b)}.$

Subtracting 1 gives: $S = \frac{1}{(1-a)(1-b)} - 1.$

Step 4: Simplify the result

Simplify the expression: $S = \frac{1 - (1-a)(1-b)}{(1-a)(1-b)}.$

Expanding the numerator: $1 - (1-a)(1-b) = 1 - (1 - a - b + ab) = a + b - ab.$

Thus: $S = \frac{a + b - ab}{(1-a)(1-b)}.$

Final Answer:

$\boxed{\frac{a + b - ab}{(1-a)(1-b)}}$

Would you like me to explain further or solve similar problems?

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5 Related Questions:

1. How can the formula for the sum of an infinite geometric series be derived?
2. What happens to the series if $|a| \geq 1$ or $|b| \geq 1$?
3. How is the double summation property used in evaluating infinite series?
4. Can you generalize this problem to sequences involving more variables (e.g., $c, d, \dots$)?
5. What are the conditions for convergence of an infinite series in general?

Tip:

Always verify the convergence conditions for any infinite series before attempting to find its sum!