Convergence & divergence of geometric series

  1. Geometric Series Overview:
  1. Convergence of Geometric Series
    Show that the following series are convergent and find its sum:
    1. n=013n\sum_{n=0}^{\infty} \frac{1}{3^n}
    2. n=1[(58)n1+(1+3n7n)] \sum_{n=1}^{\infty} [(-\frac{5}{8})^{n-1}+(\frac{1+3^n}{7^n})]
    3. n=04n+2234n\sum_{n=0}^{\infty}4^{n+2}2^{3-4n}
    4. n=042(n+2)53n1\sum_{n=0}^{\infty} \frac{4^{2(n+2)}}{5^{3n-1}}
  2. Divergence of Geometric Series
    Show that the following series are divergent:
    1. n=03n12n \sum_{n=0}^{\infty} \frac{3^{n-1}}{2^n}
    2. n=03n+223n\sum_{n=0}^{\infty}3^{n+2}2^{3-n}
    3. n=0[(14)n+(32)n2n]\sum_{n=0}^{\infty}[(\frac{1}{4})^n+(\frac{3}{2})^n2^n]
Topic Notes
In this section, we will take a look at the convergence and divergence of geometric series. We've learned about geometric sequences in high school, but in this lesson we will formally introduce it as a series and determine if the series is divergent or convergent. For the first few questions we will determine the convergence of the series, and then find the sum. For the last few questions, we will determine the divergence of the geometric series, and show that the sum of the series is infinity.
Formulas for Geometric Series:

n=0arn=a1r\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r} if -1 < rr < 1
n=1arn1=a1r\sum_{n=1}^{\infty}ar^{n-1}=\frac{a}{1-r} if -1 < rr < 1
If -1 < rr < 1, then the geometric series converges. Otherwise, the series diverges.