"Determine whether the infinite geometric series converges or..." solved and explained

tibukooinm

Answered question

2022-03-26

Determine whether the infinite geometric series converges or diverges. If it converges, find its sum.

\sum_{k = 1}^{\infty} \frac{1}{3} 4^{k = 1}

Answer & Explanation

armejantm925

Beginner2022-03-27Added 20 answers

Step 1
$\sum_{k = 1}^{\infty} \frac{1}{3} 4^{k - 1}$
Concept Used:
Series Ratio test:
$\sum_{n = 1}^{\infty} a_{n}$
$L = lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} |$
If $L < 1$ , series converges.
If $L > 1$ , series diverges.
$L = 1$ , test fail.
Step 2
Calculation:
$\sum_{k = 1}^{\infty} \frac{1}{3} 4^{k - 1}$
$= \frac{1}{3} \sum_{k = 1}^{\infty} 4^{k - 1}$
$\sum_{k = 1}^{\infty} 4^{k - 1}$ , Apply series ratio test
$L = lim_{k \to \infty} | \frac{a_{k + 1}}{a_{k}} |$
$L = \lim_{k \to \infty} | \frac{4^{k - 1 + 1}}{4^{k - 1}} |$
$L = \lim_{k \to \infty} | \frac{4^{k}}{4^{k - 1}} |$
$L = \lim_{k \to \infty} | 4^{k - k + 1} |$
$L = 4^{1}$
$L = 4 > 1$
Hence
$\sum_{k = 1}^{\infty} 4^{k - 1}$ is diverges.
$\frac{1}{3} \sum_{k = 1}^{\infty} 4^{k - 1}$ is also diverges.
$\sum_{k = 1}^{\infty} \frac{1}{3} 4^{k - 1}$ diverges.

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