find lim_(n→+∞) Σ_(k=1) ^n sin((1/(k+n)))


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Question Number 78263 by msup trace by abdo last updated on 15/Jan/20

$f i n d {l i m}_{n \to + \infty} \sum_{k = 1}^{n} s i n (\frac{1}{k + n})$
Commented by jagoll last updated on 15/Jan/20

$\underset{k = 1}{\sum^{n}} \lim_{n \to + \infty} \sin (\frac{1}{k + n}) = 0$
Commented by msup trace by abdo last updated on 15/Jan/20

$s h o w y o u r w o r k s i r . i t s i n s u f f i c i e n t . . .$
	Answered by mind is power last updated on 15/Jan/20
	$lemna ∀x>0 x−(x^3 /6)≤sin(x)≤x proff since sin(x)=Σ_(k≥0) (((−1)^k x^(2k+1) )/((2k+1)!))=x−(x^3 /6)+Σ_(k≥2) (((−1)^k x^(2k+1) )/((2k+1)!))≥x−(x^3 /6) sin(x)=x+Σ_(k≥1) (((−1)^k x^(2k+1) )/((2k+1)!))<x ⇒ x−(x^3 /6)<sin(x)<x ∀k∈[1,n] ∀n∈N^∗ we have (1/(k+n))−(1/6)((1/(k+n)))^3 <sin((1/(k+n)))<(1/(k+n)) ⇒Σ_(k=1) ^n {((1/(k+n)))−(1/(6(k+n)^3 ))}<Σ_(k=1) ^n sin((1/(k+n)))<Σ_(k=1) ^n (1/(k+n)) lim_(n→∞) Σ_(k=1) ^n (1/(n+k))=lim_(n→∞) (1/n)Σ_(k=1) ^n (1/(1+(k/n)))=∫_0 ^1 (dx/(1+x))=ln(2) Σ_(k=1) ^n (1/((n+k)^3 ))≤(1/n^2 ).Σ_(k=1) ^n (1/(n+k))≤((ln(2))/n^2 )→0 lim_(n→∞) {Σ_(k=1) ^n (1/(k+n))−(1/6)Σ_(k=1) ^n (1/((n+k)^3 ))}<lim_(n→∞) Σsin((1/(k+n)))<lim_(n→∞) Σ(1/(n+k)) ln(2)≤lim_(n→∞) Σ_(k=1) ^n sin((1/(n+k)))≤ln(2) lim_(n→∞) Σsin((1/(n+k)))=ln(2)$
	$lemna$ $\forall x > 0$ $x - \frac{x^{3}}{6} ⩽ \sin (x) ⩽ x$ $proff$ $since$ $\sin (x) = \sum_{k ⩾ 0} \frac{{(- 1)}^{k} x^{2 k + 1}}{(2 k + 1)!} = x - \frac{x^{3}}{6} + \sum_{k ⩾ 2} \frac{{(- 1)}^{k} x^{2 k + 1}}{(2 k + 1)!} ⩾ x - \frac{x^{3}}{6}$ $\sin (x) = x + \sum_{k ⩾ 1} \frac{{(- 1)}^{k} x^{2 k + 1}}{(2 k + 1)!} < x$ $\Rightarrow x - \frac{x^{3}}{6} < \sin (x) < x$ $\forall k \in [1, n] \forall n \in N^{*} we have$ $\frac{1}{k + n} - \frac{1}{6} {(\frac{1}{k + n})}^{3} < \sin (\frac{1}{k + n}) < \frac{1}{k + n}$ $\Rightarrow \underset{k = 1}{\sum^{n}} {(\frac{1}{k + n}) - \frac{1}{6 {(k + n)}^{3}}} < \underset{k = 1}{\sum^{n}} \sin (\frac{1}{k + n}) < \underset{k = 1}{\sum^{n}} \frac{1}{k + n}$ $\lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \frac{1}{n + k} = \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} \frac{1}{1 + \frac{k}{n}} = \int_{0}^{1} \frac{dx}{1 + x} = \ln (2)$ $\underset{k = 1}{\sum^{n}} \frac{1}{{(n + k)}^{3}} ⩽ \frac{1}{n^{2}} . \underset{k = 1}{\sum^{n}} \frac{1}{n + k} ⩽ \frac{\ln (2)}{n^{2}} \to 0$ $\lim_{n \to \infty} {\underset{k = 1}{\sum^{n}} \frac{1}{k + n} - \frac{1}{6} \underset{k = 1}{\sum^{n}} \frac{1}{{(n + k)}^{3}}} < \lim_{n \to \infty} Σ \sin (\frac{1}{k + n}) < \lim_{n \to \infty} Σ \frac{1}{n + k}$ $\ln (2) ⩽ \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \sin (\frac{1}{n + k}) ⩽ \ln (2)$ $\lim_{n \to \infty} Σ \sin (\frac{1}{n + k}) = \ln (2)$
	Commented by msup trace by abdo last updated on 15/Jan/20

	$t h a n k y o u s i r .$
	Commented by mind is power last updated on 15/Jan/20

	$y^{'} re welcom$
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