Find whether the series diverges and its sum: \sum_{n=1}^\infty(-1)^{n+1}\frac{3}{5^n}

xnlghtmarexgo 2021-11-23 Answered
Find whether the series diverges and its sum:
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\left(-{1}\right)}^{{{n}+{1}}}{\frac{{{3}}}{{{5}^{{n}}}}}\)

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Expert Answer

Florence Pittman
Answered 2021-11-24 Author has 1185 answers

Notice that
\(\displaystyle{\left(-{1}\right)}^{{{n}+{1}}}{\frac{{{3}}}{{{5}^{{n}}}}}=-{3}{\frac{{{\left(-{1}\right)}^{{n}}}}{{{5}^{{n}}}}}=-{3}{\left({\frac{{-{1}}}{{{5}}}}\right)}^{{n}}\)
Since \(\displaystyle{\sum_{{{k}={1}}}^{\infty}}{a}{r}^{{k}}={\frac{{{a}{r}}}{{{1}-{r}}}}\) (if \(\displaystyle{\left|{r}\right|}{<}{1}\))
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}-{3}{\left({\frac{{-{1}}}{{{5}}}}\right)}^{{n}}={\frac{{-{3}\cdot{\frac{{-{1}}}{{{5}}}}}}{{{1}-{\frac{{-{1}}}{{{5}}}}}}}={\frac{{{\frac{{{3}}}{{{5}}}}}}{{{\frac{{{6}}}{{{5}}}}}}}={\frac{{{1}}}{{{2}}}}\)
and the sum converges because \(\displaystyle{\left|{\frac{{-{1}}}{{{5}}}}\right|}={\frac{{{1}}}{{{5}}}}{<}{1}\)

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pseudoenergy34
Answered 2021-11-25 Author has 1276 answers
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