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Question Number 69338 by Rasheed.Sindhi last updated on 22/Sep/19

Commented by Prithwish sen last updated on 22/Sep/19

$it is the series of$ $1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + . . . . . .$ $\tan^{- 1} x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + . . . . . .$ $Now put x = 1$
	Answered by mind is power last updated on 22/Sep/19
	$Σ_(n=1) ^(+∞) ((sin(na))/n)=ImΣ_(n≥1) (e^(ian) /n)=Im(Σ_(n≥1) (((e^(ia) )^n )/n))=Im(−ln(1−e^(ia) )) =Im(−ln(e^(i(a/2)) (e^(−((ia)/2)) −e^((ia)/2) ))=Im{(−ln(e^(i(a/2)) )−ln(−2isin((a/2)))} =im(−i(a/2)−ln(−2i)−ln(sin((a/2))) =im(−i(a/2)−ln(2)+i(π/2)−ln((a/2))−ln(sin((a/2))) =((π−a)/2) for a=(π/2) we get =(π/4)$
	$\sum_{n = 1}^{+ \infty} \frac{s i n (n a)}{n} = I m \sum_{n ⩾ 1} \frac{e^{i a n}}{n} = I m (\sum_{n ⩾ 1} \frac{{(e^{i a})}^{n}}{n}) = I m (- l n (1 - e^{i a}))$ $= I m (- l n (e^{i \frac{a}{2}} (e^{- \frac{i a}{2}} - e^{\frac{i a}{2}})) = I m {(- l n (e^{i \frac{a}{2}}) - l n (- 2 i s i n (\frac{a}{2}))}$ $= i m (- i \frac{a}{2} - l n (- 2 i) - l n (s i n (\frac{a}{2}))$ $= i m (- i \frac{a}{2} - l n (2) + i \frac{π}{2} - l n (\frac{a}{2}) - l n (s i n (\frac{a}{2}))$ $= \frac{π - a}{2}$ $f o r a = \frac{π}{2} w e g e t = \frac{π}{4}$
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