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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

	$c . Let S = a + a r + {a r}^{2} + . . . + {a r}^{n - 1} . . . . (i)$ $now r S = r a + {a r}^{2} + {a r}^{3} + . . . + {a r}^{n} . . . . . (i i)$ $S - r S = a - {a r}^{n}$ $\Rightarrow S (1 - r) = a (1 - r^{n})$ $\Rightarrow S = \frac{a (1 - r^{n})}{1 - r} for ∣ r ∣ < 1$
	Answered by Rio Michael last updated on 05/Aug/20
	$b. let u_n = {n^2 } u_(n+1) = (n+1)^2 = n^2 + 2n + 1 u_(n+1) −u_n = n^2 +2n + 1 −n^(2 ) = 2n + 1 ∀ n ∈ N, 2n + 1 > 0 hence u_(n+1) >u_n thus u_n is monotonically increasing. let n <<1 but n ∈ N , n^2 → 0 hence n is monotonically decreasing$
	$b . let u_{n} = {n^{2}}$ $u_{n + 1} = {(n + 1)}^{2} = n^{2} + 2 n + 1$ $u_{n + 1} - u_{n} = n^{2} + 2 n + 1 - n^{2} = 2 n + 1$ $\forall n \in N, 2 n + 1 > 0 hence u_{n + 1} > u_{n}$ $thus u_{n} is monotonically increasing .$ $let n << 1 but n \in N, n^{2} \to 0$ $hence n is monotonically decreasing$
	Answered by Rio Michael last updated on 05/Aug/20
	$a. let u_n = {(1 + (5/n))^n }_(n=1) ^∞ If lim_(n→∞) u_n = L , L ∈ R ⇒ u_n converges otherwise it is divergent. lim_(n→∞) (1 + (5/n))^n = e^5 you need proof? here is it let (5/n) = (1/m) as n →∞, m → ∞ ⇒ 5m = n lim_(n→∞) (1 + (5/n))^n = lim_(m→∞) [(1 + (1/m))^m ]^5 = e^5 ⇒ u_n is convergnt$
	$a . let u_{n} = {{(1 + \frac{5}{n})}^{n}}_{n = 1}^{\infty}$ $If \lim_{n \to \infty} u_{n} = L, L \in R \Rightarrow u_{n} converges otherwise$ $it is divergent .$ $\lim_{n \to \infty} {(1 + \frac{5}{n})}^{n} = e^{5}$ $you need proof ? here is it$ $let \frac{5}{n} = \frac{1}{m} as n \to \infty, m \to \infty$ $\Rightarrow 5 m = n$ $\lim_{n \to \infty} {(1 + \frac{5}{n})}^{n} = \lim_{m \to \infty} {[{(1 + \frac{1}{m})}^{m}]}^{5} = e^{5}$ $\Rightarrow u_{n} is convergnt$
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