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Question Number 119979 by huotpat last updated on 28/Oct/20

Commented by huotpat last updated on 28/Oct/20

$N w h e n n \to + \infty i s n o t x \to + \infty$
	Answered by mathmax by abdo last updated on 28/Oct/20

	$let f (x) = \frac{{(x + 1)}^{n + 1} - (x + 1)}{x^{3}} - \frac{n}{x^{2}} - \frac{n (n + 1)}{2 x} \Rightarrow$ $f (x) = \frac{{(x + 1)}^{n + 1} - (x + 1) - nx - 2^{- 1} n (n + 1) x^{2}}{x^{3}} we knowthat$ ${(1 + x)}^{α} = 1 + \frac{α}{1!} x + \frac{α (α - 1)}{2!} x^{2} + \frac{α (α - 1) (α - 2)}{3!} x^{3} + o (x^{4}) \Rightarrow$ ${(1 + x)}^{n + 1} = 1 + (n + 1) x + \frac{(n + 1) n}{2} x^{2} + \frac{(n + 1) n (n - 1)}{6} x^{3} + o (x^{4}) \Rightarrow$ ${(x + 1)}^{n + 1} - (x + 1) - nx - 2^{- 1} n (n + 1) x^{2} \sim$ $1 + (n + 1) x + \frac{n (n + 1)}{2} x^{2} + \frac{(n - 1) n (n + 1)}{6} x^{3} - x - 1 - nx - \frac{1}{2} n (n + 1) x^{2}$ $= \frac{n (n - 1) (n + 1)}{6} x^{3} \Rightarrow f (x) \sim \frac{n (n - 1) (n + 1)}{6} \Rightarrow$ $\lim_{x \to 0} f (x) = \frac{n (n - 1) (n + 1)}{6}$
	Answered by Dwaipayan Shikari last updated on 28/Oct/20

	$\lim_{n \to + \infty} n [\underset{p}{\prod^{n}} (\underset{k = 1}{\sum^{p}} \frac{1}{k} - \frac{1}{k + 1})]$ $\lim_{n \to + \infty} n [\underset{p}{\prod^{n}} (1 - \frac{1}{p + 1})]$ $\lim_{n \to + \infty} n [\underset{p}{\prod^{n}} \frac{p}{p + 1}] = n . \frac{1}{2} . \frac{2}{3} . \frac{3}{4} . . . . . . \frac{1}{n} . \frac{n}{n + 1}$ $\lim_{n \to \infty} = \frac{n}{1 + n} = \frac{1}{1 + \frac{1}{n}} = 1$
	Answered by mathmax by abdo last updated on 28/Oct/20
	$let g(x) =((x+2x^2 +3x^3 +...+nx^n )/(x−1))−((n(n+1))/(2(x−1))) (x→1) we have 1+x +x^2 +...+x^n =((x^(n+1) −1)/(x−1)) (x≠1) by d erivation we get 1+2x+...+nx^(n−1) =(dx/dx)(((x^(n+1) −1)/(x−1))) =((nx^(n+1) −(n+1)x^n +1)/((x−1)^2 )) ⇒ x +2x^2 +3x^3 +....+nx^n =((nx^(n+2) −(n+1)x^(n+1) +x)/((x−1)^2 )) ⇒ g(x)=((nx^(n+2) −(n+1)x^(n+1) +x)/((x−1)^2 ))−((n(n+1))/(2(x−1))) ⇒ g(x)=((2nx^(n+2) −2(n+1)x^(n+1) +2x)/(2(x−1)^2 ))−((n(n+1)(x−1))/(2(x−1)^2 )) =((2nx^(n+2) −2(n+1)x^(n+1) +2x−n(n+1)(x−1))/(2(x−1)^2 )) let u(x)=2nx^(n+2) −2(n+1)x^(n+1) +2x−n(n+1)(x−1) v(x)=2(x−1)^2 ⇒ u^′ (x)=2n(n+2)x^(n+1) −2(n+1)^2 x^n +2−n(n+1) u^((2)) (x) =2n(n+2)(n+1)x^n −2n(n+1)^2 x^(n−1) ⇒ u^((2)) (1) =2n(n+2)(n+1)−2n(n+1)^2 =2n(n+1){n+2−n−1) =2n(n+1) v(x)=2(x−1)^2 ⇒v^′ (x)=4(x−1) ⇒v^((2)) (x)=4 ⇒ lim_(x→1) g(x) =lim_(x→1) ((u^((2)) (x))/(v^((2)) (x))) =((2n(n+1))/4) =((n(n+1))/2)$
	$let g (x) = \frac{x + {2 x}^{2} + {3 x}^{3} + . . . + {nx}^{n}}{x - 1} - \frac{n (n + 1)}{2 (x - 1)} (x \to 1)$ $we have 1 + x + x^{2} + . . . + x^{n} = \frac{x^{n + 1} - 1}{x - 1} (x \neq 1) by d erivation we get$ $1 + 2 x + . . . + {nx}^{n - 1} = \frac{dx}{dx} (\frac{x^{n + 1} - 1}{x - 1}) = \frac{{nx}^{n + 1} - (n + 1) x^{n} + 1}{{(x - 1)}^{2}} \Rightarrow$ $x + {2 x}^{2} + {3 x}^{3} + . . . . + {nx}^{n} = \frac{{nx}^{n + 2} - (n + 1) x^{n + 1} + x}{{(x - 1)}^{2}} \Rightarrow$ $g (x) = \frac{{nx}^{n + 2} - (n + 1) x^{n + 1} + x}{{(x - 1)}^{2}} - \frac{n (n + 1)}{2 (x - 1)} \Rightarrow$ $g (x) = \frac{{2 nx}^{n + 2} - 2 (n + 1) x^{n + 1} + 2 x}{2 {(x - 1)}^{2}} - \frac{n (n + 1) (x - 1)}{2 {(x - 1)}^{2}}$ $= \frac{{2 nx}^{n + 2} - 2 (n + 1) x^{n + 1} + 2 x - n (n + 1) (x - 1)}{2 {(x - 1)}^{2}}$ $let u (x) = {2 nx}^{n + 2} - 2 (n + 1) x^{n + 1} + 2 x - n (n + 1) (x - 1)$ $v (x) = 2 {(x - 1)}^{2} \Rightarrow$ $u^{^{'}} (x) = 2 n (n + 2) x^{n + 1} - 2 {(n + 1)}^{2} x^{n} + 2 - n (n + 1)$ $u^{(2)} (x) = 2 n (n + 2) (n + 1) x^{n} - 2 n {(n + 1)}^{2} x^{n - 1} \Rightarrow$ $u^{(2)} (1) = 2 n (n + 2) (n + 1) - 2 n {(n + 1)}^{2}$ $= 2 n (n + 1) {n + 2 - n - 1) = 2 n (n + 1)$ $v (x) = 2 {(x - 1)}^{2} \Rightarrow v^{^{'}} (x) = 4 (x - 1) \Rightarrow v^{(2)} (x) = 4 \Rightarrow$ $\lim_{x \to 1} g (x) = \lim_{x \to 1} \frac{u^{(2)} (x)}{v^{(2)} (x)} = \frac{2 n (n + 1)}{4} = \frac{n (n + 1)}{2}$
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