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

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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Florence Pittman

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
I also need the answer for this question