# Use the formula for the sum of a geometric series to find the sum, or state that the series diverges. frac{25}{9}+frac{5}{3}+1+frac{3}{5}+frac{9}{25}+frac{27}{125}+...

Question
Series
Use the formula for the sum of a geometric series to find the sum, or state that the series diverges.
$$\displaystyle{\frac{{{25}}}{{{9}}}}+{\frac{{{5}}}{{{3}}}}+{1}+{\frac{{{3}}}{{{5}}}}+{\frac{{{9}}}{{{25}}}}+{\frac{{{27}}}{{{125}}}}+\ldots$$

2021-03-10
The given geometric series is,
$$\displaystyle{\frac{{{25}}}{{{9}}}}+{\frac{{{5}}}{{{3}}}}+{1}+{\frac{{{3}}}{{{5}}}}+{\frac{{{9}}}{{{25}}}}+{\frac{{{27}}}{{{125}}}}+\ldots$$
The first term is, $$\displaystyle{a}_{{1}}={\frac{{{25}}}{{{9}}}}$$
Determine the common ratio.
$$\displaystyle{d}={\frac{{{a}_{{2}}}}{{{a}_{{1}}}}}$$
$$\displaystyle={\frac{{{\left({\frac{{{5}}}{{{3}}}}\right)}}}{{{\left({\frac{{{25}}}{{{9}}}}\right)}}}}$$
$$\displaystyle={\left({\frac{{{5}}}{{{3}}}}\right)}{\left({\frac{{{9}}}{{{25}}}}\right)}$$
$$\displaystyle={\frac{{{3}}}{{{5}}}}$$
Since r
The sum of the geometric series is,
$$\displaystyle{S}_{{n}}={\frac{{{a}{\left({1}-{r}^{{n}}\right)}}}{{{1}-{r}}}}$$
Substitute the values.
$$\displaystyle{S}_{{n}}={\left({\frac{{{25}}}{{{9}}}}\right)}{\frac{{{\left({1}-{\left({\frac{{{3}}}{{{5}}}}\right)}^{{n}}\right)}}}{{{1}-{\left({\frac{{{3}}}{{{5}}}}\right)}}}}$$
$$\displaystyle={\left({\frac{{{25}}}{{{9}}}}\right)}{\frac{{{\left({\frac{{{5}^{{n}}-{3}^{{n}}}}{{{5}^{{n}}}}}\right)}}}{{{\left({\frac{{{5}-{3}}}{{{5}}}}\right)}}}}$$
$$\displaystyle{S}_{{n}}={\left({\frac{{{25}}}{{{9}}}}\right)}{\left({\frac{{{5}^{{n}}-{3}^{{n}}}}{{{5}^{{n}}}}}\right)}{\left({\frac{{{5}}}{{{2}}}}\right)}$$
$$\displaystyle{S}_{{n}}={\left({\frac{{{5}^{{3}}}}{{{18}}}}\right)}{\left({\frac{{{5}^{{n}}-{3}^{{n}}}}{{{5}^{{n}}}}}\right)}$$
$$\displaystyle={\left({\frac{{{1}}}{{{18}}}}\right)}{\left({\frac{{{5}^{{n}}-{3}^{{n}}}}{{{5}^{{{n}-{3}}}}}}\right)}$$
Sum of the series to infinite terms is,
$$\displaystyle{S}_{{\infty}}={\frac{{{a}}}{{{1}-{r}}}}$$
$$\displaystyle={\frac{{{\left({\frac{{{25}}}{{{9}}}}\right)}}}{{{1}-{\left({\frac{{{3}}}{{{5}}}}\right)}}}}$$
$$\displaystyle={\frac{{{\left({\frac{{{25}}}{{{9}}}}\right)}}}{{{\left({\frac{{{5}-{3}}}{{{5}}}}\right)}}}}$$
$$\displaystyle={\left({\frac{{{25}}}{{{9}}}}\right)}{\left({\frac{{{5}}}{{{2}}}}\right)}$$
$$\displaystyle={\frac{{{125}}}{{{18}}}}$$

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