...nice mathematics... prove that::: i:: Σ_(n=1) ^∞ (1/(sinh^2 (πn))) =(1/6) −(1/(2π)) ✓ ii:: Σ_(n=1) ^∞ (n/(e^(2πn) −1))=(1/(24)) −(1/(8π)) ✓✓ iii::Σ_(n=1) ^∞ (1/( nsinh(πn))) =(π/(12))−((ln(2))/4) ✓✓✓ .... M..n..july..1970 ....


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Question Number 114996 by mnjuly1970 last updated on 22/Sep/20

$. . . n i c e m a t h e m a t i c s . . .$ $p r o v e t h a t :::$ $i :: \underset{n = 1}{\sum^{\infty}} \frac{1}{{s i n h}^{2} (π n)} = \frac{1}{6} - \frac{1}{2 π} ✓$ $i i :: \underset{n = 1}{\sum^{\infty}} \frac{n}{e^{2 π n} - 1} = \frac{1}{24} - \frac{1}{8 π} ✓ ✓$ $i i i :: \underset{n = 1}{\sum^{\infty}} \frac{1}{n s i n h (π n)} = \frac{π}{12} - \frac{l n (2)}{4} ✓ ✓ ✓$ $. . . . M . . n . . j u l y . . 1970 . . . .$
Commented by maths mind last updated on 24/Sep/20

$i w i l l p o s t e a l l m y w o r c k l a t e r$ $j u s t i i) i f o u n d 2 (\frac{1}{24} - \frac{1}{8 π}) . . .$
	Answered by Olaf last updated on 23/Sep/20

	$i ::$ $I_{n} = \int_{n}^{n + 1} \frac{d x}{\sinh^{2} (π x)}$ $I_{n} = \int_{n}^{n + 1} (\coth^{2} (π x) - 1) d x$ $I_{n} = [{(- \frac{1}{π} \coth (π x)]}_{n}^{n + 1} = \frac{1}{π} [\coth (π n) - \coth (π (n + 1))]$ $I_{n} = \frac{1}{π} [\frac{\cosh (π n)}{\sinh (π n)} - \frac{\cosh (π (n + 1))}{\sinh (π (n + 1))}]$ $I_{n} = \frac{1}{π} \frac{\sinh (π (n + 1)) \cosh (π n) - \cosh (π (n + 1) + \cosh (π n)}{\sinh (π n) \sinh (π (n + 1))}$ $I_{n} = \frac{1}{π} [\frac{\sinh (π (n + 1) - π n]}{\sinh (π n) \sinh (π (n + 1))}]$ $I_{n} = \frac{1}{π} [\frac{\sinh (π)}{\sinh (π n) \sinh (π (n + 1))}]$ $\frac{π}{\sinh (π)} I_{n} = \frac{1}{\sinh (π n) \sinh (π (n + 1))}$ $\frac{π}{\sinh (π)} I_{n + 1} ⩽ \frac{1}{\sinh^{2} (π (n + 1))} ⩽ \frac{π}{\sinh (π)} I_{n}$ $\frac{π}{\sinh (π)} \underset{n = 1}{\sum^{\infty}} I_{n + 1} ⩽ \underset{n = 1}{\sum^{\infty}} \frac{1}{\sinh^{2} (π (n + 1))} ⩽ \underset{n = 1}{\sum^{\infty}} \frac{π}{\sinh (π)} I_{n}$ $\frac{π}{\sinh (π)} \underset{n = 2}{\sum^{\infty}} I_{n} ⩽ \underset{n = 1}{\sum^{\infty}} \frac{1}{\sinh^{2} (π n)} - \frac{1}{\sinh^{2} (π)} ⩽ \underset{n = 1}{\sum^{\infty}} \frac{π}{\sinh (π)} I_{n}$ $\frac{π}{\sinh (π)} {[- \frac{1}{π} \coth (π x)]}_{2}^{\infty} ⩽ \underset{n = 1}{\sum^{\infty}} \frac{1}{\sinh^{2} (π n)} - \frac{1}{\sinh^{2} (π)} ⩽ \frac{π}{\sinh (π)} {[- \frac{1}{π} \coth (π x)]}_{1}^{\infty}$ $\frac{1}{\sinh (π)} [\coth (2 π) - 1] ⩽ \underset{n = 1}{\sum^{\infty}} \frac{1}{\sinh^{2} (π n)} - \frac{1}{\sinh^{2} (π)} ⩽ \frac{1}{\sinh (π)} [\coth (π) - 1]$ $\frac{1}{\sinh (π)} [\coth (2 π) - 1 + \frac{1}{\sinh (π)}] ⩽ \underset{n = 1}{\sum^{\infty}} \frac{1}{\sinh^{2} (π n)} ⩽ \frac{1}{\sinh (π)} {[\coth (π) - 1 + \frac{1}{\sinh (π)}]}_{1}^{\infty}$ $\frac{\coth (2 π) \sinh (π) - \sinh (π) + 1}{\sinh^{2} (π)} ⩽ \underset{n = 1}{\sum^{\infty}} \frac{1}{\sinh^{2} (π n)} ⩽ [\frac{\cosh (π) - \sinh (π) + 1}{\sinh^{2} (π)}]$ $I tried but may be it is not the good way .$
	Answered by maths mind last updated on 24/Sep/20
	$ii) Σ(n/(e^(2πn) −1)) let f(z)=(z/(e^(2πz) −1)) holomorphic cunction over C−{iZ} it has rvable singularity at origine lim_(z→0) (z/(e^(2πz) −1))=(1/(2π)) we can use Σ_(n≥0) f(n)=(1/(2π))+Σ_(n≥1) f(n) Σ_(n≥0) f(n)=∫_0 ^∞ f(x)dx+((f(0))/2)+i∫_0 ^∞ ((f(it)−f(−it))/(e^(2πt) −1))dt Abel −plana formula f(it)−f(−it) =((it)/(e^(2iπt) −1))+((it)/(e^(−2iπt) −1)) =−it⇒i∫_0 ^∞ ((f(it)−f(−it))/(e^(2πt) −1))dt=∫_0 ^∞ (x/(e^(2πx) −1))dx Σ_(n≥0) f(n)_(=S) =(1/(2π))+Σ_(n≥1) f(n)=(1/(2π))+Σ_(n≥1) f(n)=(1/(4π))+2∫_0 ^∞ ((xdx)/(e^(2πx) −1)) =2∫_0 ^∞ ((xdx)/(e^(2πx) −1)),2πx=t⇒dx=(dt/(2π))⇒ S=(1/(4π))−(1/(2π))+(1/(2π^2 ))∫_0 ^∞ (t/(e^t −1))dt=−(1/(4π))+(1/(2π^2 ))ζ(2)Γ(2)=−(1/(4π))+(1/(12)). Σ(n/(e^(2πn) −1))=(1/(12))−(1/(4π))$
	$i i)$ $Σ \frac{n}{e^{2 π n} - 1}$ $l e t f (z) = \frac{z}{e^{2 π z} - 1} h o l o m o r p h i c c u n c t i o n o v e r C - {i Z}$ $i t h a s r v a b l e s i n g u l a r i t y a t o r i g i n e$ $\lim_{z \to 0} \frac{z}{e^{2 π z} - 1} = \frac{1}{2 π}$ $w e c a n u s e$ $\sum_{n ⩾ 0} f (n) = \frac{1}{2 π} + \sum_{n ⩾ 1} f (n)$ $\sum_{n ⩾ 0} f (n) = \int_{0}^{\infty} f (x) d x + \frac{f (0)}{2} + i \int_{0}^{\infty} \frac{f (i t) - f (- i t)}{e^{2 π t} - 1} d t A b e l - p l a n a f o r m u l a$ $f (i t) - f (- i t)$ $= \frac{i t}{e^{2 i π t} - 1} + \frac{i t}{e^{- 2 i π t} - 1} = - i t \Rightarrow i \int_{0}^{\infty} \frac{f (i t) - f (- i t)}{e^{2 π t} - 1} d t = \int_{0}^{\infty} \frac{x}{e^{2 π x} - 1} d x$ $Missing \left or extra \right$ $= 2 \int_{0}^{\infty} \frac{x d x}{e^{2 π x} - 1}, 2 π x = t \Rightarrow d x = \frac{d t}{2 π} \Rightarrow$ $S = \frac{1}{4 π} - \frac{1}{2 π} + \frac{1}{2 π^{2}} \int_{0}^{\infty} \frac{t}{e^{t} - 1} d t = - \frac{1}{4 π} + \frac{1}{2 π^{2}} ζ (2) Γ (2) = - \frac{1}{4 π} + \frac{1}{12} .$ $Σ \frac{n}{e^{2 π n} - 1} = \frac{1}{12} - \frac{1}{4 π}$
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