calculate Σ_(n=1) ^∞ ((cos(nθ))/n^2 ) and Σ


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Question Number 45518 by maxmathsup by imad last updated on 14/Oct/18

$c a l c u l a t e \sum_{n = 1}^{\infty} \frac{c o s (n θ)}{n^{2}} a n d \sum_{n = 1}^{\infty} \frac{s i n (n θ)}{n^{2}}$
Commented by maxmathsup by imad last updated on 23/Oct/18
$we have proved that Σ_(n=1) ^∞ ((cos(nx))/n) =ln(sin((x/2))) ⇒ ∫_0 ^x (Σ_(n=1) ^∞ ((cos(nt))/n))dt =∫_0 ^x ln(sin((t/2)))dt ⇒ Σ_(n=1) ^∞ (1/n)[(1/n)sin(nt)]_0 ^x = ∫_0 ^x ln(sin((t/2)))dt⇒ Σ_(n=1) ^∞ ((sin(nx))/n^2 ) = ∫_0 ^x ln(sin((t/2)))dt=_((t/2)=u ) 2 ∫_0 ^(x/2) ln(sin(u))du also we have Σ_(n=1) ^∞ ((sin(nx))/n) =((π−x)/2) ⇒ ∫_0 ^x (Σ_(n=1) ^∞ ((sin(nt))/n))dt =∫_0 ^x ((π−t)/2)dt ⇒ Σ_(n=1) ^∞ (1/n) ∫_0 ^x sin(nt)dt =((πx)/2) −(1/2) (x^2 /2) ⇒ Σ_(n=1) ^∞ (1/n)[−(1/n)cos(nt)]_0 ^x =((πx)/2) −(x^2 /4) ⇒ Σ_(n=1) ^∞ (1/n^2 ){1−cos(nx)} =((πx)/2) −(x^2 /4) ⇒ Σ_(n=1) ^∞ ((cos(nx))/n^2 ) =(π^2 /6) −((πx)/2) +(x^2 /4) .$
$w e h a v e p r o v e d t h a t \sum_{n = 1}^{\infty} \frac{c o s (n x)}{n} = l n (s i n (\frac{x}{2})) \Rightarrow$ $\int_{0}^{x} (\sum_{n = 1}^{\infty} \frac{c o s (n t)}{n}) d t = \int_{0}^{x} l n (s i n (\frac{t}{2})) d t \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{1}{n} {[\frac{1}{n} s i n (n t)]}_{0}^{x} = \int_{0}^{x} l n (s i n (\frac{t}{2})) d t \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{s i n (n x)}{n^{2}} = \int_{0}^{x} l n (s i n (\frac{t}{2})) d t =_{\frac{t}{2} = u} 2 \int_{0}^{\frac{x}{2}} l n (s i n (u)) d u a l s o w e h a v e$ $\sum_{n = 1}^{\infty} \frac{s i n (n x)}{n} = \frac{π - x}{2} \Rightarrow \int_{0}^{x} (\sum_{n = 1}^{\infty} \frac{s i n (n t)}{n}) d t = \int_{0}^{x} \frac{π - t}{2} d t \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{1}{n} \int_{0}^{x} s i n (n t) d t = \frac{π x}{2} - \frac{1}{2} \frac{x^{2}}{2} \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{1}{n} {[- \frac{1}{n} c o s (n t)]}_{0}^{x} = \frac{π x}{2} - \frac{x^{2}}{4} \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} {1 - c o s (n x)} = \frac{π x}{2} - \frac{x^{2}}{4} \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{c o s (n x)}{n^{2}} = \frac{π^{2}}{6} - \frac{π x}{2} + \frac{x^{2}}{4} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18

	$P_{n} + {i Q}_{n} = \frac{c o s n θ}{n^{2}} + i \frac{s i n n θ}{n^{2}} = \frac{e^{i n θ}}{n^{2}}$ $T_{n} = P + {i Q}_{n} = \frac{e^{i n θ}}{n^{2}}$ $S_{n} = \underset{n = 1}{\sum^{\infty}} T_{n}$ $S_{n} = \frac{e^{i θ}}{1^{2}} + \frac{e^{i 2 θ}}{2^{2}} + \frac{e^{i 3 θ}}{3^{2}} + . . . \infty$ $S_{n} = \frac{x}{1^{2}} + \frac{x^{2}}{2^{2}} + \frac{x^{3}}{3^{2}} + . . . \infty$ $w a i t p l z . . .$
	Answered by Meritguide1234 last updated on 15/Oct/18

	Commented by maxmathsup by imad last updated on 21/Oct/18

	$y o u r a n s w e r i s n o t c o r r e c t s i r b e c a u s e ∣ \frac{s i n θ}{n^{2}} ∣⩽ \frac{1}{n^{2}} a n d Σ \frac{1}{n^{2}} c o n v e r g e s . . .$
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