Expert answers to "Determined the convergence of this alternating series. The..."

Bobbie Comstock

Answered question

2021-12-22

Determined the convergence of this alternating series. The series is $\sum_{i = 1}^{\infty} 2 (\frac{- 1}{4})^{(n - 1)}$

Elaine Verrett

Beginner2021-12-23Added 41 answers

The other submitter had the general idea. The series

\sum_{i = 1}^{\infty} 2 {(\frac{- 1}{4})}^{(n - 1)}

is a geometric series of the form

\sum_{n = 1}^{\infty} a r^{n - 1} = a + a r + a r^{2} + a r^{3} + \dots

where the first term is a and the common ratio is r. This geometric series always converges if

| r | < 1

, where in this case the sum is given by:

S_{\infty} = \frac{a}{1 - r}

Comparing coefficients, it can be seen that

a = 2

and

r = \frac{- 1}{4}

Therefore, since

| r | = \frac{1}{4} < 1

, the sum of the series is given by:

S_{\infty} = \frac{a}{1 - r} = \frac{2}{1 - \frac{- 1}{4}} = \frac{8}{5}

Kindlein6h

Beginner2021-12-24Added 27 answers

\sum_{n = 1}^{\infty} {(- \frac{1}{4})}^{n - 1} = \sum_{n = 1}^{\infty} x^{n - 1} {|_{{- \frac{1}{4}}} = \frac{1}{1 - x} |}_{- \frac{1}{4}} = \frac{4}{5}

Since (

| x | < 1

)

\sum_{n = 0}^{\infty} = \frac{1}{1 - x}

user_27qwe

Skilled2021-12-30Added 375 answers

Hint:
$\sum_{i = 0}^{n} q^{i} = \frac{1 - q^{n + 1}}{1 - q}$

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