Calc6.35

From Example Problems
Jump to: navigation, search

Discuss the convergence or divergence of the series \sum _{{n=7}}^{{\infty }}\left[\left(-{\frac  {4}{7}}\right)^{{n+3}}+\left({\frac  {1}{3}}\right)^{{n-2}}\right].

This problem is a mix of every problem we’ve done thus far. Let’s split this up into two sums and reindex.

S=\sum _{{n=7}}^{{\infty }}\left[\left(-{\frac  {4}{7}}\right)^{{n+3}}+\left({\frac  {1}{3}}\right)^{{n-2}}\right]=\sum _{{n=10}}^{{\infty }}\left(-{\frac  {4}{7}}\right)^{n}+\sum _{{n=5}}^{{\infty }}\left({\frac  {1}{3}}\right)^{n}\,

So, we have two different geometric series, both of which converge so their sum also converges and converges to

S=\sum _{{n=0}}^{{\infty }}\left(-{\frac  {4}{7}}\right)^{n}-\sum _{{n=0}}^{{9}}\left(-{\frac  {4}{7}}\right)^{n}+\sum _{{n=0}}^{{\infty }}\left({\frac  {1}{3}}\right)^{n}-\sum _{{n=0}}^{{4}}\left({\frac  {1}{3}}\right)^{n}\,

={\frac  {1}{1+{\frac  {4}{7}}}}-{\frac  {1-\left(-{\frac  {4}{7}}\right)^{{10}}}{1+{\frac  {4}{7}}}}+{\frac  {1}{1-{\frac  {1}{3}}}}-{\frac  {1-\left({\frac  {1}{3}}\right)^{5}}{1-{\frac  {1}{3}}}}={\frac  {4}{11}}\left({\frac  {4}{7}}\right)^{9}+{\frac  {1}{162}}={\frac  {613758989}{71910127674}}\,


Main Page : Calculus