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
asked 2021-03-09
Use the formula for the sum of a geometric series to find the sum, or state that the series diverges.
\(\displaystyle{\frac{{{25}}}{{{9}}}}+{\frac{{{5}}}{{{3}}}}+{1}+{\frac{{{3}}}{{{5}}}}+{\frac{{{9}}}{{{25}}}}+{\frac{{{27}}}{{{125}}}}+\ldots\)

Answers (1)

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,
\(\displaystyle{S}_{{\infty}}={\frac{{{a}}}{{{1}-{r}}}}\)
\(\displaystyle={\frac{{{\left({\frac{{{25}}}{{{9}}}}\right)}}}{{{1}-{\left({\frac{{{3}}}{{{5}}}}\right)}}}}\)
\(\displaystyle={\frac{{{\left({\frac{{{25}}}{{{9}}}}\right)}}}{{{\left({\frac{{{5}-{3}}}{{{5}}}}\right)}}}}\)
\(\displaystyle={\left({\frac{{{25}}}{{{9}}}}\right)}{\left({\frac{{{5}}}{{{2}}}}\right)}\)
\(\displaystyle={\frac{{{125}}}{{{18}}}}\)
0

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