Question

Use the formula for the sum of a geometric series to find the sum. sum_{n=4}^infty(-frac49)^n

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asked 2020-10-20
Use the formula for the sum of a geometric series to find the sum.
\(\sum_{n=4}^\infty(-\frac49)^n\)

Answers (1)

2020-10-21
Given series:
\(\sum_{n=4}^\infty(-\frac49)^n\)
For the given series the ratio of (n+1)-th term to n-th term is constant which is \(-\frac{4}{9}\)
Hence, the terms of the given series are in geometric progression.
Write the given series in standard geometric series form.
That is,
\(\sum_{n=4}^\infty(-\frac49)^n=(-\frac49)^{-4}+(-\frac{4}{9})^{-3}+(-\frac49)^{-2}+(-\frac49)^{-1}+\sum_{n=0}^\infty(-\frac49)^n\)
\(\sum_{n=4}^\infty(-\frac49)^n=\frac{4365}{256}+\sum_{n=0}^\infty(-\frac49)^n\)
The general geometric series,
\(\sum_{n=0}^\infty ar^n\), where a is the first term and r is the common ratio
converges to \(\frac{a}{1-r}\), for -1 Otherwise it diverges.
Compare the series \(\sum_{n=0}^\infty(-\frac49)^n\) with the general geometric series, to get
\(a=1,r=-\frac{4}{9}\int(-1,1)\)
Hence, \(\sum_{n=0}^\infty(-\frac49)^n=\frac{1}{1-(-\frac49)}\)
\(=\frac{1}{1+\frac49}\)
\(=\frac{1}{(\frac{13}{9})}\)
\(=\frac{9}{13}\)
Plug \(\sum_{n=0}^\infty(-\frac49)^n=\frac{9}{13}\) in equation (1), to get
\(\sum_{n=-4}^\infty(-\frac49)^n=\frac{4365}{256}+\frac{9}{13}\)
\(=\frac{59049}{3328}\)
\(\approx17.7431\)
Thus,
\(\sum_{n=-4}^\infty(-\frac49)^n=17.7431\)
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