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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

$$\mathrm{Exactly} \\ $$$$\:{a}\:+\:{ar}\:+\:{ar}^{\mathrm{2}} \:+...+{ar}^{{n}−\mathrm{1}} \:=\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{ar}^{{n}−\mathrm{1}} \:=\:\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}} \\ $$$$\mathrm{for}\:\mathrm{convergence}\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{S}_{{n}} \:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{ar}^{{n}−\mathrm{1}} \\ $$$$\Rightarrow\:{S}_{\infty} \:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}}\right]\: \\ $$$$\mathrm{if}\:\mid{r}\mid\:<\:\mathrm{1}\:\:\mathrm{then}\:\mathrm{as}\:{n}\:\rightarrow\:\infty\:,\:{r}^{{n}} \:\rightarrow\:\mathrm{0} \\ $$$$\Rightarrow\:\:{S}_{\infty} \:=\:\frac{{a}}{\mathrm{1}−{r}}. \\ $$$$\mathrm{notice}\:\mathrm{if}\:\mid{r}\mid\:>\:\mathrm{1}\:,\:\:{r}^{{n}} \rightarrow\infty\:\mathrm{as}\:{r}\:\rightarrow\:\infty\: \\ $$$$\Rightarrow\:{S}_{\infty} \:=\:\infty\:\mathrm{thus}\:\mathrm{for}\:\mathrm{convergence},\:\:\:\mathrm{of}\:{S}_{{n}} ,\:\:\mid{r}\mid\:<\:\mathrm{1} \\ $$

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