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,

\(\sum_{n=-4}^\infty(-\frac49)^n=17.7431\)

\(\sum_{n=4}^\infty(-\frac49)^n\)

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,

\(\sum_{n=-4}^\infty(-\frac49)^n=17.7431\)