find-n-0-1-3n-1-1-3-1-6-1-9-

Question Number 178793 by infinityaction last updated on 21/Oct/22

find \underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n)!} = 1 + \frac{1}{3!} + \frac{1}{6!} + \frac{1}{9!} + \dots

Answered by aleks041103 last updated on 21/Oct/22

w = e^{\frac{2 π i}{3}} \Rightarrow w^{2} = w^{*}, w^{3} = 1

e^{w} = \underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n)!} + \frac{w}{(3 n + 1)!} + \frac{w^{2}}{(3 n + 2)!} = S_{0} + {w S}_{1} + w^{2} S_{2}

e^{w^{2}} = \underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n)!} + \frac{w^{2}}{(3 n + 1)!} + \frac{w}{(3 n + 2)!} = S_{0} + w^{2} S_{1} + {w S}_{2}

e^{1} = S_{0} + S_{1} + S_{2}

e^{w} + e^{w^{2}} - 2 e = (w + w^{2} - 2) (S_{1} + S_{2}) = - 3 (S_{1} + S_{2})

\Rightarrow S_{0} = e - (S_{1} + S_{2}) = e + \frac{(e x p (w) + e x p (w^{2}) - 2 e)}{3} =

= \frac{e^{1} + e^{w} + e^{w^{2}}}{3} = S_{0}

w = e^{\frac{2 π i}{3}} = - \frac{1}{2} + i \frac{\sqrt{3}}{2}

w^{2} = e^{- \frac{2 π i}{3}} = - \frac{1}{2} - i \frac{\sqrt{3}}{2}

\Rightarrow e^{w} = \frac{1}{\sqrt{e}} e^{i \sqrt{3} / 2}, e^{w^{2}} = \frac{1}{\sqrt{e}} e^{- i \sqrt{3} / 2}

\Rightarrow e^{w} + e^{w^{2}} = \frac{2}{\sqrt{e}} c o s (\frac{\sqrt{3}}{2})

\Rightarrow \underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n)!} = \frac{e}{3} + \frac{2}{3 \sqrt{e}} c o s (\sqrt{3} / 2)

Commented by aleks041103 last updated on 21/Oct/22

Commented by Tawa11 last updated on 22/Oct/22

Great sir

Commented by infinityaction last updated on 22/Oct/22

t h a n k s s i r

Answered by ARUNG_Brandon_MBU last updated on 22/Oct/22

S = \underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n)!} = 1 + \frac{1}{3!} + \frac{1}{6!} + \frac{1}{9!} + \cdot \cdot \cdot

= (1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \frac{1}{7!} + \frac{1}{8!} + \frac{1}{9!} + \cdot \cdot \cdot) - (\frac{1}{1!} + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{5!} + \frac{1}{7!} + \frac{1}{8!} + \cdot \cdot \cdot)

= e - (\frac{1}{1!} + \frac{1}{4!} + \frac{1}{7!} + \cdot \cdot \cdot) - (\frac{1}{2!} + \frac{1}{5!} + \frac{1}{8!} + \cdot \cdot \cdot) = e - \underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n + 1)!} - \underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n + 2)!}

\Rightarrow e = \underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n)!} + \underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n + 1)!} + \underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n + 2)!}

e^{x} = \underset{n = 0}{\sum^{\infty}} \frac{x^{3 n}}{(3 n)!} + \underset{n = 0}{\sum^{\infty}} \frac{x^{3 n + 1}}{(3 n + 1)!} + \underset{n = 0}{\sum^{\infty}} \frac{x^{3 n + 2}}{(3 n + 2)} \Rightarrow e^{x} = f (x) + f^{'} (x) + f^{″} (x)

\underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{(3 n)!} is solution to this equation with f (0) = 1 and f^{'} (0) = 0

y_{g h} : r^{2} + r + 1 = 0 \Rightarrow r = - \frac{1}{2} \pm i \frac{\sqrt{3}}{2} \Rightarrow y_{g h} = e^{- \frac{x}{2}} (α \cos (\frac{\sqrt{3}}{2} x) + β \sin (\frac{\sqrt{3}}{2} x))

y_{p} = {k e}^{x} = y_{p}^{'} = y_{p}^{″} \Rightarrow e^{x} = {k e}^{x} + {k e}^{x} + {k e}^{x} \Rightarrow k = \frac{1}{3}

f (x) = y_{g h} + y_{p} = e^{- \frac{x}{2}} (α \cos (\frac{\sqrt{3}}{2} x) + β \sin (\frac{\sqrt{3}}{2} x)) + \frac{e^{x}}{3}, f (0) = 1 \Rightarrow α + \frac{1}{3} = 1 \Rightarrow α = \frac{2}{3}

f^{'} (x) = e^{- \frac{x}{2}} (\frac{\sqrt{3}}{2} β \cos (\frac{\sqrt{3}}{2} x) - \frac{\sqrt{3}}{2} α \sin (\frac{\sqrt{3}}{2} x)) - \frac{1}{2} e^{- \frac{x}{2}} (α \cos (\frac{\sqrt{3}}{2} x) + β \sin (\frac{\sqrt{3}}{2} x)) + \frac{e^{x}}{3}

f^{'} (0) = 0 \Rightarrow \frac{\sqrt{3}}{2} β - \frac{α}{2} + \frac{1}{3} = \frac{\sqrt{3}}{2} β = 0 \Rightarrow β = 0 \Rightarrow f (x) = e^{- \frac{x}{2}} (\frac{2}{3} \cos (\frac{\sqrt{3}}{2} x)) + \frac{e^{x}}{3}

\underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n)!} = f (1) = \frac{2}{3 \sqrt{e}} \cos (\frac{\sqrt{3}}{2}) + \frac{e}{3} = \frac{1}{3} (e + \frac{2}{\sqrt{e}} \cos (\frac{\sqrt{3}}{2}))

Commented by infinityaction last updated on 22/Oct/22

t h a n k s