What is the sum of the series [math: (−1/3)0+(−1/3)1+(−1/3)2+(−1/3)3+…](-1/3)^0+(-1/3)^1+(-1/3)^2+(-1/3)^3+\ldots Remember to enter your answer in lowest terms.

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BlackTom AI · Accepted Answer

The problem presents an infinite geometric series with terms formed as (-1/3)^n for n = 0, 1, 2, ... .

First, identify the key parameters of a geometric series: the first term a, and the common ratio r between successive terms. Here, the first term is a = (-1/3)^0 = 1, and the common ratio is r = -1/3 since each term multiplies by -1/3 to get the next term.

Next, check convergence. An infinite geometric series converges if |r| < 1. Since |r| = |-1/3| = 1/3 < 1, the series does converge and we can use the standard sum formula for an infinite geometric series.

Apply the sum formula S = a / (1 - r). Substituting a = 1 and r = -1/3, we get S = 1 / (1 - (-1/3)) = 1 / (1 + 1/3) = 1 / (4/3).

Compute the division: dividing by a fraction is the same as multiplying by its reciprocal, so S = 1 × (3/4) = 3/4.

Finally, this result is already in lowest terms because 3/4 cannot be reduced further. The steps above show how the sum is derived from the geometric-series framework, using the first term and the common ratio and confirming convergence before applying the formula.

What is the sum of the series [math: (−1/3)0+(−1/3)1+(−1/3)2+(−1/3)3+…](-1/3)^0+(-1/3)^1+(-1/3)^2+(-1/3)^3+\ldots Remember to enter your answer in lowest terms.

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