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Section 10.5 : Special Series

5. Determine if the series converges or diverges. If the series converges give its value.

\[\sum\limits_{n = 1}^\infty {\frac{{{5^{n + 1}}}}{{{7^{n - 2}}}}} \]

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Given that all three of the special series we looked at in this section are all pretty distinct it is hopefully clear that this is a geometric series.

Show Step 2

Let’s also notice that the initial value of the index is \(n = 1\) and so we can put this into the form,

\[\sum\limits_{n = 1}^\infty {a\,{r^{n - 1}}} \]

At that point we’ll be able to determine if it converges or diverges and the value of the series if it does happen to converge.

As noted above we need the two exponents to be \(n - 1\). This is an easy “fix” if we note that using basic exponent properties we can write each term as follows,

\[{5^{n + 1}} = {5^{n - 1}}{5^2}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{7^{n - 2}} = {7^{n - 1}}{7^{ - 1}}\]

With these two rewrites the series becomes,

\[\sum\limits_{n = 1}^\infty {\frac{{{5^{n + 1}}}}{{{7^{n - 2}}}}} = \sum\limits_{n = 1}^\infty {\frac{{{5^{n - 1}}{5^2}}}{{{7^{n - 1}}{7^{ - 1}}}}} = \sum\limits_{n = 1}^\infty {\left( {25} \right)\left( 7 \right)\frac{{{5^{n - 1}}}}{{{7^{n - 1}}}}} = \sum\limits_{n = 1}^\infty {175{{\left( {\frac{5}{7}} \right)}^{n - 1}}} \] Show Step 3

With the series in “proper” form we can see that \(a = 175\) and \(r = \frac{5}{7}\). Therefore, because we can clearly see that \(\left| r \right| = \frac{5}{7} < 1\), the series will converge and its value is,

\[\sum\limits_{n = 1}^\infty {\frac{{{5^{n + 1}}}}{{{7^{n - 2}}}}} = \sum\limits_{n = 1}^\infty {175{{\left( {\frac{5}{7}} \right)}^{n - 1}}} = \frac{{175}}{{1 - \frac{5}{7}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{1225}}{2}}}\]