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Question Number 106479 by mohammad17 last updated on 05/Aug/20

Answered by Rio Michael last updated on 05/Aug/20

c. Let S = a + ar + ar^2 + ...+ ar^(n−1) ....(i) now rS = ra + ar^2 + ar^3 +...+ar^n .....(ii) S−rS = a−ar^n ⇒ S(1−r) = a(1−r^n ) ⇒ S = ((a(1−r^n ))/(1−r)) for ∣r∣ < 1

$$\mathrm{c}.\:\mathrm{Let}\:\:{S}\:=\:{a}\:+\:{ar}\:+\:{ar}^{\mathrm{2}} \:+\:...+\:{ar}^{{n}−\mathrm{1}} \:....\left({i}\right) \\ $$$$\:\:\:\mathrm{now}\:{rS}\:=\:{ra}\:+\:{ar}^{\mathrm{2}} \:+\:{ar}^{\mathrm{3}} \:+...+{ar}^{{n}} .....\left({ii}\right) \\ $$$${S}−{rS}\:=\:{a}−{ar}^{{n}} \\ $$$$\Rightarrow\:{S}\left(\mathrm{1}−{r}\right)\:=\:{a}\left(\mathrm{1}−{r}^{{n}} \right) \\ $$$$\Rightarrow\:{S}\:=\:\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}}\:\:\:\mathrm{for}\:\:\mid{r}\mid\:<\:\mathrm{1} \\ $$

Answered by Rio Michael last updated on 05/Aug/20

b. let u_n = {n^2 } u_(n+1) = (n+1)^2 = n^2 + 2n + 1 u_(n+1) −u_n = n^2 +2n + 1 −n^(2 ) = 2n + 1 ∀ n ∈ N, 2n + 1 > 0 hence u_(n+1) >u_n thus u_n is monotonically increasing. let n <<1 but n ∈ N , n^2 → 0 hence n is monotonically decreasing

$$\mathrm{b}.\:\:\mathrm{let}\:{u}_{{n}} \:=\:\left\{{n}^{\mathrm{2}} \right\} \\ $$$${u}_{{n}+\mathrm{1}} \:=\:\left({n}+\mathrm{1}\right)^{\mathrm{2}} \:=\:{n}^{\mathrm{2}} \:+\:\mathrm{2}{n}\:+\:\mathrm{1} \\ $$$${u}_{{n}+\mathrm{1}} −{u}_{{n}} \:=\:{n}^{\mathrm{2}} \:+\mathrm{2}{n}\:+\:\mathrm{1}\:−{n}^{\mathrm{2}\:} \:=\:\mathrm{2}{n}\:+\:\mathrm{1} \\ $$$$\forall\:{n}\:\in\:\mathbb{N},\:\:\mathrm{2}{n}\:+\:\mathrm{1}\:>\:\mathrm{0}\:\:\mathrm{hence}\:\:{u}_{{n}+\mathrm{1}} \:>{u}_{{n}} \\ $$$$\mathrm{thus}\:{u}_{{n}} \:\mathrm{is}\:\mathrm{monotonically}\:\mathrm{increasing}. \\ $$$$\mathrm{let}\:{n}\:<<\mathrm{1}\:\mathrm{but}\:{n}\:\in\:\mathbb{N}\:,\:{n}^{\mathrm{2}} \:\rightarrow\:\mathrm{0} \\ $$$$\mathrm{hence}\:{n}\:\mathrm{is}\:\mathrm{monotonically}\:\mathrm{decreasing} \\ $$

Answered by Rio Michael last updated on 05/Aug/20

a. let u_n = {(1 + (5/n))^n }_(n=1) ^∞ If lim_(n→∞) u_n = L , L ∈ R ⇒ u_n converges otherwise it is divergent. lim_(n→∞) (1 + (5/n))^n = e^5 you need proof? here is it let (5/n) = (1/m) as n →∞, m → ∞ ⇒ 5m = n lim_(n→∞) (1 + (5/n))^n = lim_(m→∞) [(1 + (1/m))^m ]^5 = e^5 ⇒ u_n is convergnt

$$\mathrm{a}.\:\:\mathrm{let}\:{u}_{{n}} \:=\:\left\{\left(\mathrm{1}\:+\:\frac{\mathrm{5}}{{n}}\right)^{{n}} \right\}_{{n}=\mathrm{1}} ^{\infty} \\ $$$$\mathrm{If}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{u}_{{n}} \:=\:{L}\:\:,\:{L}\:\in\:\mathbb{R}\:\Rightarrow\:{u}_{{n}} \:\:\mathrm{converges}\:\mathrm{otherwise} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{divergent}. \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}\:+\:\frac{\mathrm{5}}{{n}}\right)^{{n}} \:=\:{e}^{\mathrm{5}} \\ $$$$\mathrm{you}\:\mathrm{need}\:\mathrm{proof}?\:\mathrm{here}\:\mathrm{is}\:\mathrm{it} \\ $$$$\mathrm{let}\:\frac{\mathrm{5}}{{n}}\:=\:\frac{\mathrm{1}}{{m}}\:\mathrm{as}\:{n}\:\rightarrow\infty,\:{m}\:\rightarrow\:\infty \\ $$$$\Rightarrow\:\mathrm{5}{m}\:=\:{n} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}\:+\:\frac{\mathrm{5}}{{n}}\right)^{{n}} \:=\:\underset{{m}\rightarrow\infty} {\mathrm{lim}}\:\left[\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{{m}}\right)^{{m}} \right]^{\mathrm{5}} =\:{e}^{\mathrm{5}} \\ $$$$\Rightarrow\:{u}_{{n}} \:\mathrm{is}\:\mathrm{convergnt} \\ $$

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