lim_(x→∞) (1/x)Σ_(r=1) ^x cos(((rπ)/(2x))) x∈N


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Question Number 158334 by alcohol last updated on 02/Nov/21

$\lim_{x \to \infty} \frac{1}{x} \underset{r = 1}{\sum^{x}} c o s (\frac{r π}{2 x})$ $x \in N$
Commented by aleks041103 last updated on 02/Nov/21

$T h e u p p e r b o u n d o f t h e s u m m a t i o n$ $c a n o n l y b e a n a t u r a l n u m b e r, w h i l e$ $x i s n o t . . . P r o b a b l y t h e r e i s a m i s t a k e!$
Commented by alcohol last updated on 02/Nov/21

$x \in N$
Commented by puissant last updated on 03/Nov/21
$lim_(x→∞) (1/x)Σ_(r=1) ^x cos(((rπ)/(2x))) = ∫_0 ^1 cos(((πa)/2))da = (2/π){sin(((πa)/2))}_0 ^1 = (2/π)$
$\lim_{x \to \infty} \frac{1}{x} \underset{r = 1}{\sum^{x}} c o s (\frac{r π}{2 x}) = \int_{0}^{1} c o s (\frac{π a}{2}) d a$ $= \frac{2}{π} {s i n (\frac{π a}{2})}_{0}^{1} = \frac{2}{π}$
	Answered by aleks041103 last updated on 02/Nov/21

	$\underset{r = 1}{\sum^{x}} c o s (\frac{r π}{2 x}) = R e (\underset{r = 1}{\sum^{x}} e^{\frac{i r π}{2 x}}) = R e (\underset{r = 1}{\sum^{x}} {(e^{\frac{i π}{2 x}})}^{r}) =$ $= R e (e^{\frac{i π}{2 x}} \frac{{(e^{\frac{i π}{2 x}})}^{x} - 1}{e^{\frac{i π}{2 x}} - 1}) =$ $= R e (\frac{e^{\frac{i π}{2 x}}}{e^{\frac{i π}{2 x}} - 1}) R e ({(e^{\frac{i π}{2 x}})}^{x} - 1) - I m (\frac{e^{\frac{i π}{2 x}}}{e^{\frac{i π}{2 x}} - 1}) I m ({(e^{\frac{i π}{2 x}})}^{x} - 1)$ $\frac{e^{\frac{i π}{2 x}}}{e^{\frac{i π}{2 x}} - 1} = 1 + \frac{1}{e^{\frac{i π}{2 x}} - 1} = 1 + \frac{e^{- \frac{i π}{2 x}} - 1}{(e^{\frac{i π}{2 x}} - 1) (e^{- \frac{i π}{2 x}} - 1)} =$ $= 1 + \frac{e^{- \frac{i π}{2 x}} - 1}{1 + 1 - (e^{\frac{i π}{2 x}} + e^{- \frac{i π}{2 x}})} = 1 + \frac{e^{- \frac{i π}{2 x}} - 1}{2 (1 - c o s (\frac{π}{2 x}))} = \frac{1}{2} (1 - i c t g (\frac{π}{4 x}))$ $\Rightarrow R e (e^{\frac{i π}{2 x}} \frac{{(e^{\frac{i π}{2 x}})}^{x} - 1}{e^{\frac{i π}{2 x}} - 1}) = R e (\frac{e^{\frac{i π}{2 x}}}{e^{\frac{i π}{2 x}} - 1} (i - 1))$ $= \frac{1}{2} (- 1) - \frac{1}{2} c t g (\frac{π}{4 x}) = - \frac{1}{2} (1 + c t g (\frac{π}{4 x})) . . .$
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