nice-mathematics-prove-that-i-n-1-1-sinh-2-pin-1-6-1-2pi-ii-n-1-n-e-2pin-1-1-24-

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

\dots n i c e m a t h e m a t i c s \dots

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} ✓ ✓ ✓

\dots . M . . n . . j u l y . . 1970 \dots .

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}) \dots

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

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}