(1/1^2 )+(1/2^2 )+(1/3^2 )+(1/4^2 )+(1/6^2 )+.....∞=?


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Question Number 99853 by Dwaipayan Shikari last updated on 23/Jun/20

$\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \frac{1}{6^{2}} + . . . . . \infty = ?$
	Answered by smridha last updated on 23/Jun/20

	$\underset{n = 1}{\sum^{\infty}} n^{- 2} = ζ (2) = \frac{π^{2}}{6}$
	Commented by Dwaipayan Shikari last updated on 23/Jun/20

	$T h a n k s f o r g i v i n g t h e a n s w e r . C a n y o u d e s c r i b e i t . A s i a m a s t u d e n t i w a n t t o l e a r n i t .$
	Commented by abdomathmax last updated on 23/Jun/20

	$can you show a way for this mr mridha . . .$
	Commented by smridha last updated on 23/Jun/20

	$o k k k i c a n$
	Commented by smridha last updated on 23/Jun/20

	$B_{2 n} = {(- 1)}^{n - 1} \frac{2 (2 n)!}{{(2 π)}^{2 n}} ζ (2 n)$ $t h i s i s t h e r e l a t i o n {b e t}^{n} B e r n o u l l i$ $n u m b e r a n d R e i m a n n z e t a f^{n}$ $n o w p u t n = 1 y o u g e t . .$ $B_{2} = {(- 1)}^{1 - 1} \frac{2.2!}{{(2 π)}^{2}} ζ (2)$ $w e k n o w B_{2} = \frac{1}{6} p u t t i n g i t y o u g e t$ $ζ (2) = \frac{π^{2}}{6} .$
	Commented by smridha last updated on 23/Jun/20

	$a n o t h e r w a y :$ $s i n x = x \underset{n = 1}{\prod^{\infty}} (1 - \frac{x^{2}}{n^{2} π^{2}})$ $\Rightarrow \frac{s i n (x)}{x} = (1 - \frac{x^{2}}{1^{2} π^{2}}) (1 - \frac{x^{2}}{2^{2} π^{2}}) (1 - \frac{x^{2}}{3^{2} π^{2}}) . . . . . . . .$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (i)$ $w e k n o w : s i n x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - . . . . . .$ $s o \frac{s i n (x)}{x} = 1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - . . . . . . . .$ $n o w c o m p a r e t h e x^{2} t e r m s b o t h$ $s i d e s o f t h e {e q}^{n} (i) w e w i l l g e t$ $\frac{1}{1^{2} π^{2}} + \frac{1}{2^{2} π^{2}} + \frac{1}{3^{2} π^{2}} + . . . . . = \frac{1}{3!}$ $s o f i n a l l y w e g e t :$ $\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . . \infty = ζ (2) = \frac{π^{2}}{6} .$
	Commented by I want to learn more last updated on 23/Jun/20

	$sir what of ζ (3) ? ? ?$
	Commented by abdomathmax last updated on 23/Jun/20

	$ξ (3) = \sum_{n = 1}^{\infty} \frac{1}{n^{3}} = 1 + \frac{1}{8} + \frac{1}{27} + . . . .$ $ξ (3) \sim 1, 2$
	Commented by I want to learn more last updated on 23/Jun/20

	$Sir, please full workings .$
	Commented by smridha last updated on 23/Jun/20

	$b y E u l e r - M a c l a u r i n i n t e g r a t i o n$ $f o r m u l a . . .$ $ζ (3) = \frac{1}{2} + \frac{1}{2} + \underset{p = 1}{\sum^{\infty}} \frac{(2 p + 1) B_{2 p}}{2 x^{2 p + 2}} + r e m a i d e r .$ $ζ (3) \approx 1.202057 . .$
	Commented by mathmax by abdo last updated on 23/Jun/20

	$i think no elementary function to calculate the exact value$
	Commented by smridha last updated on 24/Jun/20

	$t h e r e f o r e I u s e d \approx i n s t e a d o f$ $= .$
	Answered by abdomathmax last updated on 23/Jun/20
	$let f(x)=1 2π periodic odd let developp f at fourier serie f(x) =Σ_(n=1) ^∞ a_n sin(nx) a_n =(2/T)∫_([T]) sin(nx)dx =(1/π)∫_(−π) ^π sin(nx)dx =(2/π) ∫_0 ^π sin(nx)dx =−(2/π)[(1/n)cos(nx)]_0 ^π =−(2/(nπ)){ (−1)^n −1) ⇒ f(x) =Σ_(n=1) ^(∞ ) (2/(nπ))(1−(−1)^n )sin(nx) =(2/π) Σ_(n=0) ^∞ (2/(2n+1))sin(2n+1)x =1 ⇒ (4/π) Σ_(n=0) ^∞ ((sin(2n+1)x)/((2n+1))) =1 ⇒ Σ_(n=0) ^∞ ((sin(2n+1)x)/(2n+1)) =(π/4) let integrate ⇒ −Σ_(n=0) ^∞ ((cos(2n+1)x)/((2n+1)^2 )) =((πx)/4) +c x=0 ⇒−Σ_(n=0) ^∞ (1/((2n+1)^2 )) =c ⇒ ((πx)/4) =−Σ_(n=0) ^∞ ((cos(2n+1)x)/((2n+1)^2 ))+Σ_(n=0) ^∞ (1/((2n+1)^2 )) x=π ⇒(π^2 /4) =Σ_(n=0) ^∞ (1/((2n+1)^2 )) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(π^2 /8) we have Σ_(n=1) ^∞ (1/n^2 ) =(1/4)Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒(3/4)Σ_(n=1) ^∞ (1/n^2 ) =(π^2 /8) ⇒ Σ_(n=1) ^∞ (1/n^2 ) =(4/3)×(π^2 /8) Σ_(n=1) ^∞ (1/n^2 ) =(π^2 /6)$
	$let f (x) = 1 2 π periodic odd let developp f at$ $fourier serie f (x) = \sum_{n = 1}^{\infty} a_{n} \sin (nx)$ $a_{n} = \frac{2}{T} \int_{[T]} \sin (nx) dx = \frac{1}{π} \int_{- π}^{π} \sin (nx) dx$ $= \frac{2}{π} \int_{0}^{π} \sin (nx) dx = - \frac{2}{π} {[\frac{1}{n} \cos (nx)]}_{0}^{π}$ $= - \frac{2}{n π} {{(- 1)}^{n} - 1) \Rightarrow$ $f (x) = \sum_{n = 1}^{\infty} \frac{2}{n π} (1 - {(- 1)}^{n}) \sin (nx)$ $= \frac{2}{π} \sum_{n = 0}^{\infty} \frac{2}{2 n + 1} \sin (2 n + 1) x = 1 \Rightarrow$ $\frac{4}{π} \sum_{n = 0}^{\infty} \frac{\sin (2 n + 1) x}{(2 n + 1)} = 1 \Rightarrow$ $\sum_{n = 0}^{\infty} \frac{\sin (2 n + 1) x}{2 n + 1} = \frac{π}{4} let integrate \Rightarrow$ $- \sum_{n = 0}^{\infty} \frac{\cos (2 n + 1) x}{{(2 n + 1)}^{2}} = \frac{π x}{4} + c$ $x = 0 \Rightarrow - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = c \Rightarrow$ $\frac{π x}{4} = - \sum_{n = 0}^{\infty} \frac{\cos (2 n + 1) x}{{(2 n + 1)}^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}$ $x = π \Rightarrow \frac{π^{2}}{4} = \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow$ $\sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π^{2}}{8}$ $we have \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}$ $\Rightarrow \frac{3}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{8} \Rightarrow \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{4}{3} \times \frac{π^{2}}{8}$ $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6}$
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