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Question Number 203357 by Calculusboy last updated on 17/Jan/24

$$\mathit{F}\mathit{o}\mathit{r}\mathit{t}\mathit{h}\mathit{e}\mathit{s}\mathit{e}\mathit{r}\mathit{i}\mathit{e}\mathit{s}5-\frac{5}{2}+\frac{5}{4}-\frac{5}{8}+\cdot \cdot \cdot +\frac{{(-1)}^{\mathit{n}-1}5}{2^{\mathit{n}-1}}$$$$\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{a}\mathit{n}\mathit{e}\mathit{x}\mathit{p}\mathit{r}\mathit{e}\mathit{s}\mathit{s}\mathit{i}\mathit{o}\mathit{n}\mathit{f}\mathit{o}\mathit{r}\mathit{t}\mathit{h}\mathit{e}\mathit{s}\mathit{u}\mathit{m}\mathit{o}\mathit{f}\mathit{t}\mathit{h}\mathit{e}\mathit{f}\mathit{i}\mathit{r}\mathit{s}\mathit{t}$$$$\mathit{n}\mathit{t}\mathit{e}\mathit{r}\mathit{m}\mathit{s}.\mathit{A}\mathit{l}\mathit{s}\mathit{o}\mathit{i}\mathit{f}\mathit{t}\mathit{h}\mathit{e}\mathit{s}\mathit{e}\mathit{r}\mathit{i}\mathit{e}\mathit{s}\mathit{c}\mathit{o}\mathit{n}\mathit{v}\mathit{e}\mathit{r}\mathit{g}\mathit{e}\mathit{s},$$$$\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{t}\mathit{h}\mathit{e}\mathit{s}\mathit{u}\mathit{m}\mathit{t}\mathit{o}\mathrm{\infty }.$$$$$$$$$$

Answered by mr W last updated on 17/Jan/24

$$G.P.withq=-\frac{1}{2}$$$$S_n=\frac{5(1-{(-\frac{1}{2})}^n)}{1-(-\frac{1}{2})}=\frac{10}{3}[1-\frac{{(-1)}^n}{2^n}]$$$$\underset{n\to \mathrm{\infty }}{\lim }{S}_n=\frac{10}{3}$$

Commented by Calculusboy last updated on 17/Jan/24

$$\mathit{n}\mathit{i}\mathit{c}\mathit{e}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathit{i}\mathit{r}$$

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