lim-x-1-x-r-1-x-cos-rpi-2x-x-N-

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 \dots 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 \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})) \dots