Calc6.35

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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

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

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

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Calc6.35

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