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

Series
ANSWERED Use the formula for the sum of a geometric series to find the sum.
$$\sum_{n=4}^\infty(-\frac49)^n$$ 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,
$$\sum_{n=-4}^\infty(-\frac49)^n=17.7431$$