Question Number 126253 by mohammad17 last updated on 18/Dec/20
Answered by physicstutes last updated on 18/Dec/20
![Exactly a + ar + ar^2 +...+ar^(n−1) = Σ_(r=1) ^n ar^(n−1) = ((a(1−r^n ))/(1−r)) for convergence lim_(n→∞) S_n = lim_(n→∞) Σ_(r=1) ^n ar^(n−1) ⇒ S_∞ = lim_(n→∞) [((a(1−r^n ))/(1−r))] if ∣r∣ < 1 then as n → ∞ , r^n → 0 ⇒ S_∞ = (a/(1−r)). notice if ∣r∣ > 1 , r^n →∞ as r → ∞ ⇒ S_∞ = ∞ thus for convergence, of S_n , ∣r∣ < 1](https://www.tinkutara.com/question/Q126254.png)
$$\mathrm{Exactly} \\ $$$$\:{a}\:+\:{ar}\:+\:{ar}^{\mathrm{2}} \:+…+{ar}^{{n}−\mathrm{1}} \:=\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{ar}^{{n}−\mathrm{1}} \:=\:\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}} \\ $$$$\mathrm{for}\:\mathrm{convergence}\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{S}_{{n}} \:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{ar}^{{n}−\mathrm{1}} \\ $$$$\Rightarrow\:{S}_{\infty} \:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}}\right]\: \\ $$$$\mathrm{if}\:\mid{r}\mid\:<\:\mathrm{1}\:\:\mathrm{then}\:\mathrm{as}\:{n}\:\rightarrow\:\infty\:,\:{r}^{{n}} \:\rightarrow\:\mathrm{0} \\ $$$$\Rightarrow\:\:{S}_{\infty} \:=\:\frac{{a}}{\mathrm{1}−{r}}. \\ $$$$\mathrm{notice}\:\mathrm{if}\:\mid{r}\mid\:>\:\mathrm{1}\:,\:\:{r}^{{n}} \rightarrow\infty\:\mathrm{as}\:{r}\:\rightarrow\:\infty\: \\ $$$$\Rightarrow\:{S}_{\infty} \:=\:\infty\:\mathrm{thus}\:\mathrm{for}\:\mathrm{convergence},\:\:\:\mathrm{of}\:{S}_{{n}} ,\:\:\mid{r}\mid\:<\:\mathrm{1} \\ $$