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

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Question Number 158334 by alcohol last updated on 02/Nov/21
lim_(x→∞) (1/x)Σ_(r=1) ^x cos(((rπ)/(2x))) x∈N
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}\underset{{r}=\mathrm{1}} {\overset{{x}} {\sum}}{cos}\left(\frac{{r}\pi}{\mathrm{2}{x}}\right) \\ $$$${x}\in\mathbb{N} \\ $$
Commented by aleks041103 last updated on 02/Nov/21
The upper bound of the summation can only be a natural number, while x is not...Probably there is a mistake!
$${The}\:{upper}\:{bound}\:{of}\:{the}\:{summation} \\ $$$${can}\:{only}\:{be}\:{a}\:{natural}\:{number},\:{while} \\ $$$${x}\:{is}\:{not}…{Probably}\:{there}\:{is}\:{a}\:{mistake}! \\ $$
Commented by alcohol last updated on 02/Nov/21
x∈N
$${x}\in\mathbb{N} \\ $$
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)
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}\underset{{r}=\mathrm{1}} {\overset{{x}} {\sum}}{cos}\left(\frac{{r}\pi}{\mathrm{2}{x}}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {cos}\left(\frac{\pi{a}}{\mathrm{2}}\right){da} \\ $$$$=\:\frac{\mathrm{2}}{\pi}\left\{{sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)\right\}_{\mathrm{0}} ^{\mathrm{1}} \:=\:\frac{\mathrm{2}}{\pi} \\ $$
Answered by aleks041103 last updated on 02/Nov/21
Σ_(r=1) ^x cos(((rπ)/(2x)))=Re(Σ_(r=1) ^x e^((irπ)/(2x)) )=Re(Σ_(r=1) ^x (e^((iπ)/(2x)) )^r )= =Re(e^((iπ)/(2x)) (((e^((iπ)/(2x)) )^x −1)/(e^((iπ)/(2x)) −1)))= =Re((e^((iπ)/(2x)) /(e^((iπ)/(2x)) −1)))Re((e^((iπ)/(2x)) )^x −1)−Im((e^((iπ)/(2x)) /(e^((iπ)/(2x)) −1)))Im((e^((iπ)/(2x)) )^x −1) (e^((iπ)/(2x)) /(e^((iπ)/(2x)) −1)) =1+(1/(e^((iπ)/(2x)) −1))=1+((e^(−((iπ)/(2x))) −1)/((e^((iπ)/(2x)) −1)(e^(−((iπ)/(2x))) −1)))= =1+((e^(−((iπ)/(2x))) −1)/(1+1−(e^((iπ)/(2x)) +e^(−((iπ)/(2x))) )))=1+((e^(−((iπ)/(2x))) −1)/(2(1−cos((π/(2x))))))=(1/2)(1−ictg((π/(4x)))) ⇒Re(e^((iπ)/(2x)) (((e^((iπ)/(2x)) )^x −1)/(e^((iπ)/(2x)) −1)))=Re((e^((iπ)/(2x)) /(e^((iπ)/(2x)) −1))(i−1)) =(1/2)(−1)−(1/2)ctg((π/(4x)))=−(1/2)(1+ctg((π/(4x))))...
$$\underset{{r}=\mathrm{1}} {\overset{{x}} {\sum}}{cos}\left(\frac{{r}\pi}{\mathrm{2}{x}}\right)={Re}\left(\underset{{r}=\mathrm{1}} {\overset{{x}} {\sum}}{e}^{\frac{{ir}\pi}{\mathrm{2}{x}}} \right)={Re}\left(\underset{{r}=\mathrm{1}} {\overset{{x}} {\sum}}\left({e}^{\frac{{i}\pi}{\mathrm{2}{x}}} \right)^{{r}} \right)= \\ $$$$={Re}\left({e}^{\frac{{i}\pi}{\mathrm{2}{x}}} \frac{\left({e}^{\frac{{i}\pi}{\mathrm{2}{x}}} \right)^{{x}} −\mathrm{1}}{{e}^{\frac{{i}\pi}{\mathrm{2}{x}}} −\mathrm{1}}\right)= \\ $$$$={Re}\left(\frac{{e}^{\frac{{i}\pi}{\mathrm{2}{x}}} }{{e}^{\frac{{i}\pi}{\mathrm{2}{x}}} −\mathrm{1}}\right){Re}\left(\left({e}^{\frac{{i}\pi}{\mathrm{2}{x}}} \right)^{{x}} −\mathrm{1}\right)−{Im}\left(\frac{{e}^{\frac{{i}\pi}{\mathrm{2}{x}}} }{{e}^{\frac{{i}\pi}{\mathrm{2}{x}}} −\mathrm{1}}\right){Im}\left(\left({e}^{\frac{{i}\pi}{\mathrm{2}{x}}} \right)^{{x}} −\mathrm{1}\right) \\ $$$$\frac{{e}^{\frac{{i}\pi}{\mathrm{2}{x}}} }{{e}^{\frac{{i}\pi}{\mathrm{2}{x}}} −\mathrm{1}}\:=\mathrm{1}+\frac{\mathrm{1}}{{e}^{\frac{{i}\pi}{\mathrm{2}{x}}} −\mathrm{1}}=\mathrm{1}+\frac{{e}^{−\frac{{i}\pi}{\mathrm{2}{x}}} −\mathrm{1}}{\left({e}^{\frac{{i}\pi}{\mathrm{2}{x}}} −\mathrm{1}\right)\left({e}^{−\frac{{i}\pi}{\mathrm{2}{x}}} −\mathrm{1}\right)}= \\ $$$$=\mathrm{1}+\frac{{e}^{−\frac{{i}\pi}{\mathrm{2}{x}}} −\mathrm{1}}{\mathrm{1}+\mathrm{1}−\left({e}^{\frac{{i}\pi}{\mathrm{2}{x}}} +{e}^{−\frac{{i}\pi}{\mathrm{2}{x}}} \right)}=\mathrm{1}+\frac{{e}^{−\frac{{i}\pi}{\mathrm{2}{x}}} −\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{2}{x}}\right)\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{ictg}\left(\frac{\pi}{\mathrm{4}{x}}\right)\right) \\ $$$$\Rightarrow{Re}\left({e}^{\frac{{i}\pi}{\mathrm{2}{x}}} \frac{\left({e}^{\frac{{i}\pi}{\mathrm{2}{x}}} \right)^{{x}} −\mathrm{1}}{{e}^{\frac{{i}\pi}{\mathrm{2}{x}}} −\mathrm{1}}\right)={Re}\left(\frac{{e}^{\frac{{i}\pi}{\mathrm{2}{x}}} }{{e}^{\frac{{i}\pi}{\mathrm{2}{x}}} −\mathrm{1}}\left({i}−\mathrm{1}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ctg}\left(\frac{\pi}{\mathrm{4}{x}}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{ctg}\left(\frac{\pi}{\mathrm{4}{x}}\right)\right)… \\ $$

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